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二叉树层序遍历登场！

算法公开课

《代码随想录》算法视频公开课 (opens new window)：讲透二叉树的层序遍历 | 广度优先搜索 | LeetCode：102.二叉树的层序遍历 (opens new window)，相信结合视频再看本篇题解，更有助于大家对本题的理解。

学会二叉树的层序遍历，可以一口气打完以下十题：

• 102.二叉树的层序遍历 (opens new window)
• 107.二叉树的层次遍历II (opens new window)
• 199.二叉树的右视图 (opens new window)
• 637.二叉树的层平均值 (opens new window)
• 429.N叉树的层序遍历 (opens new window)
• 515.在每个树行中找最大值 (opens new window)
• 116.填充每个节点的下一个右侧节点指针 (opens new window)
• 117.填充每个节点的下一个右侧节点指针II (opens new window)
• 104.二叉树的最大深度 (opens new window)
• 111.二叉树的最小深度 (opens new window)

102.二叉树的层序遍历

力扣题目链接 (opens new window)

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。

思路

我们之前讲过了三篇关于二叉树的深度优先遍历的文章：

• 二叉树：前中后序递归法 (opens new window)
• 二叉树：前中后序迭代法 (opens new window)
• 二叉树：前中后序迭代方式统一写法 (opens new window)

接下来我们再来介绍二叉树的另一种遍历方式：层序遍历。

层序遍历一个二叉树。就是从左到右一层一层的去遍历二叉树。这种遍历的方式和我们之前讲过的都不太一样。

需要借用一个辅助数据结构即队列来实现，队列先进先出，符合一层一层遍历的逻辑，而用栈先进后出适合模拟深度优先遍历也就是递归的逻辑。

而这种层序遍历方式就是图论中的广度优先遍历，只不过我们应用在二叉树上。

使用队列实现二叉树广度优先遍历，动画如下：

这样就实现了层序从左到右遍历二叉树。

代码如下：这份代码也可以作为二叉树层序遍历的模板，打十个就靠它了。

c++代码如下：

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> result;
        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;
            // 这里一定要使用固定大小size，不要使用que.size()，因为que.size是不断变化的
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node->val);
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(vec);
        }
        return result;
    }
};

# 递归法
class Solution {
public:
    void order(TreeNode* cur, vector<vector<int>>& result, int depth)
    {
        if (cur == nullptr) return;
        if (result.size() == depth) result.push_back(vector<int>());
        result[depth].push_back(cur->val);
        order(cur->left, result, depth + 1);
        order(cur->right, result, depth + 1);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        int depth = 0;
        order(root, result, depth);
        return result;
    }
};

其他语言版本

Java:

// 102.二叉树的层序遍历
class Solution {
    public List<List<Integer>> resList = new ArrayList<List<Integer>>();

    public List<List<Integer>> levelOrder(TreeNode root) {
        //checkFun01(root,0);
        checkFun02(root);

        return resList;
    }

    //BFS--递归方式
    public void checkFun01(TreeNode node, Integer deep) {
        if (node == null) return;
        deep++;

        if (resList.size() < deep) {
            //当层级增加时，list的Item也增加，利用list的索引值进行层级界定
            List<Integer> item = new ArrayList<Integer>();
            resList.add(item);
        }
        resList.get(deep - 1).add(node.val);

        checkFun01(node.left, deep);
        checkFun01(node.right, deep);
    }

    //BFS--迭代方式--借助队列
    public void checkFun02(TreeNode node) {
        if (node == null) return;
        Queue<TreeNode> que = new LinkedList<TreeNode>();
        que.offer(node);

        while (!que.isEmpty()) {
            List<Integer> itemList = new ArrayList<Integer>();
            int len = que.size();

            while (len > 0) {
                TreeNode tmpNode = que.poll();
                itemList.add(tmpNode.val);

                if (tmpNode.left != null) que.offer(tmpNode.left);
                if (tmpNode.right != null) que.offer(tmpNode.right);
                len--;
            }

            resList.add(itemList);
        }

    }
}

Python:

# 利用长度法
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = collections.deque([root])
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(level)
        return result

#递归法
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []

        levels = []

        def traverse(node, level):
            if not node:
                return

            if len(levels) == level:
                levels.append([])

            levels[level].append(node.val)
            traverse(node.left, level + 1)
            traverse(node.right, level + 1)

        traverse(root, 0)
        return levels

Go:

/**
102. 二叉树的递归遍历
 */
func levelOrder(root *TreeNode) [][]int {
	arr := [][]int{}

	depth := 0

	var order func(root *TreeNode, depth int)

	order = func(root *TreeNode, depth int) {
		if root == nil {
			return
		}
		if len(arr) == depth {
			arr = append(arr, []int{})
		}
		arr[depth] = append(arr[depth], root.Val)

		order(root.Left, depth+1)
		order(root.Right, depth+1)
	}

	order(root, depth)

	return arr
}

/**
102. 二叉树的层序遍历 使用container包
 */
func levelOrder(root *TreeNode) [][]int {
    res := [][]int{}
    if root == nil{//防止为空
        return res
    }
    queue := list.New()
    queue.PushBack(root)

    var tmpArr []int

    for queue.Len() > 0 {
        length := queue.Len()               //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数）
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)    //出队列
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            tmpArr = append(tmpArr, node.Val)    //将值加入本层切片中
        }
        res = append(res, tmpArr)          //放入结果集
        tmpArr = []int{}                  //清空层的数据
    }

    return res
}

/**
  102. 二叉树的层序遍历 使用切片
*/
func levelOrder(root *TreeNode) [][]int {
     res := make([][]int, 0)
     if root == nil {
        return res
     }
     queue := make([]*TreeNode, 0)
     queue = append(queue, root)
     for len(queue) > 0 {
          size := len(queue)
          level := make([]int, 0)
          for i := 0; i < size; i++ {
               node := queue[0]
               queue = queue[1:]
               level = append(level, node.Val)
               if node.Left != nil {
				   queue = append(queue, node.Left)
               }
               if node.Right != nil {
				   queue = append(queue, node.Right)
               }
          }
          res = append(res, level)
     }
     return res
}

/**
102. 二叉树的层序遍历：使用切片模拟队列，易理解
 */
func levelOrder(root *TreeNode) (res [][]int) {
    if root == nil {
        return
    }

    curLevel := []*TreeNode{root}  // 存放当前层节点
    for len(curLevel) > 0 {
        nextLevel := []*TreeNode{}  // 准备通过当前层生成下一层
        vals := []int{}

        for _, node := range curLevel {
            vals = append(vals, node.Val) // 收集当前层的值
            // 收集下一层的节点
            if node.Left != nil {
                nextLevel = append(nextLevel, node.Left)
            }
            if node.Right != nil {
                nextLevel = append(nextLevel, node.Right)
            }
        }
        res = append(res, vals)
        curLevel = nextLevel // 将下一层变成当前层
    }

    return
}

Javascript：

var levelOrder = function(root) {
    //二叉树的层序遍历
    let res = [], queue = [];
    queue.push(root);
    if(root === null) {
        return res;
    }
    while(queue.length !== 0) {
        // 记录当前层级节点数
        let length = queue.length;
        //存放每一层的节点
        let curLevel = [];
        for(let i = 0;i < length; i++) {
            let node = queue.shift();
            curLevel.push(node.val);
            // 存放当前层下一层的节点
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
        }
        //把每一层的结果放到结果数组
        res.push(curLevel);
    }
    return res;
};

TypeScript：

function levelOrder(root: TreeNode | null): number[][] {
    let helperQueue: TreeNode[] = [];
    let res: number[][] = [];
    let tempArr: number[] = [];
    if (root !== null) helperQueue.push(root);
    let curNode: TreeNode;
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            curNode = helperQueue.shift()!;
            tempArr.push(curNode.val);
            if (curNode.left !== null) {
                helperQueue.push(curNode.left);
            }
            if (curNode.right !== null) {
                helperQueue.push(curNode.right);
            }
        }
        res.push(tempArr);
        tempArr = [];
    }
    return res;
};

Swift：

func levelOrder(_ root: TreeNode?) -> [[Int]] {
    var result = [[Int]]()
    guard let root = root else { return result }
    // 表示一层
    var queue = [root]
    while !queue.isEmpty {
        let count = queue.count
        var subarray = [Int]()
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            subarray.append(node.val)
            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
        result.append(subarray)
    }

    return result
}

Scala:

// 102.二叉树的层序遍历
object Solution {
  import scala.collection.mutable
  def levelOrder(root: TreeNode): List[List[Int]] = {
    val res = mutable.ListBuffer[List[Int]]()
    if (root == null) return res.toList
    val queue = mutable.Queue[TreeNode]() // 声明一个队列
    queue.enqueue(root) // 把根节点加入queue
    while (!queue.isEmpty) {
      val tmp = mutable.ListBuffer[Int]()
      val len = queue.size // 求出len的长度
      for (i <- 0 until len) { // 从0到当前队列长度的所有节点都加入到结果集
        val curNode = queue.dequeue()
        tmp.append(curNode.value)
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      res.append(tmp.toList)
    }
    res.toList
  }
}

Rust:

use std::cell::RefCell;
use std::rc::Rc;
use std::collections::VecDeque;
impl Solution {
    pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let mut temp = vec![];
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                temp.push(node.borrow().val);
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
            res.push(temp);
        }
        res
    }
}

C#

public IList<IList<int>> LevelOrder(TreeNode root)
{
    var res = new List<IList<int>>();
    var que = new Queue<TreeNode>();
    if (root == null) return res;
    que.Enqueue(root);
    while (que.Count != 0)
    {
        var size = que.Count;
        var vec = new List<int>();
        for (int i = 0; i < size; i++)
        {
            var cur = que.Dequeue();
            vec.Add(cur.val);
            if (cur.left != null) que.Enqueue(cur.left);
            if (cur.right != null) que.Enqueue(cur.right);
        }
        res.Add(vec);


    }
    return res;
}

此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！

107.二叉树的层次遍历 II

力扣题目链接 (opens new window)

给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

思路

相对于102.二叉树的层序遍历，就是最后把result数组反转一下就可以了。

C++代码：

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> result;
        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node->val);
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(vec);
        }
        reverse(result.begin(), result.end()); // 在这里反转一下数组即可
        return result;

    }
};

其他语言版本

Python：

class Solution:
    """二叉树层序遍历II迭代解法"""

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        queue = collections.deque([root])
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(level)
        return result[::-1]

Java：

// 107. 二叉树的层序遍历 II
public class N0107 {

    /**
     * 解法：队列，迭代。
     * 层序遍历，再翻转数组即可。
     */
    public List<List<Integer>> solution1(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        Deque<TreeNode> que = new LinkedList<>();

        if (root == null) {
            return list;
        }

        que.offerLast(root);
        while (!que.isEmpty()) {
            List<Integer> levelList = new ArrayList<>();

            int levelSize = que.size();
            for (int i = 0; i < levelSize; i++) {
                TreeNode peek = que.peekFirst();
                levelList.add(que.pollFirst().val);

                if (peek.left != null) {
                    que.offerLast(peek.left);
                }
                if (peek.right != null) {
                    que.offerLast(peek.right);
                }
            }
            list.add(levelList);
        }

        List<List<Integer>> result = new ArrayList<>();
        for (int i = list.size() - 1; i >= 0; i-- ) {
            result.add(list.get(i));
        }

        return result;
    }
}

/**
 * 思路和模板相同, 对收集答案的方式做了优化, 最后不需要反转
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        // 利用链表可以进行 O(1) 头部插入, 这样最后答案不需要再反转
        LinkedList<List<Integer>> ans = new LinkedList<>();

        Queue<TreeNode> q = new LinkedList<>();

        if (root != null) q.offer(root);

        while (!q.isEmpty()) {
            int size = q.size();

            List<Integer> temp = new ArrayList<>();

            for (int i = 0; i < size; i ++) {
                TreeNode node = q.poll();

                temp.add(node.val);

                if (node.left != null) q.offer(node.left);

                if (node.right != null) q.offer(node.right);
            }

            // 新遍历到的层插到头部, 这样就满足按照层次反序的要求
            ans.addFirst(temp);
        }

        return ans;
    }

Go:

/**
107. 二叉树的层序遍历 II
 */
func levelOrderBottom(root *TreeNode) [][]int {
    queue := list.New()
    res := [][]int{}
    if root == nil{
        return res
    }
    queue.PushBack(root)

    for queue.Len() > 0 {
        length := queue.Len()
        tmp := []int{}
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            tmp = append(tmp, node.Val)
        }
        res=append(res, tmp)
    }

    //反转结果集
    for i:=0; i<len(res)/2; i++ {
        res[i], res[len(res)-i-1] = res[len(res)-i-1], res[i]
    }

    return res
}

// 使用切片作为队列
func levelOrderBottom(root *TreeNode) [][]int {
    res := make([][]int, 0)
    if root == nil {
		return res
    }
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        level := make([]int, 0)
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            level = append(level, node.Val)
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        res = append(res, level)
    }
    l, r := 0, len(res)-1
    for l < r {
		res[l], res[r] = res[r], res[l]
		l++
		r--
    }
    return res
}

Javascript:

var levelOrderBottom = function (root) {
  let res = [],
    queue = [];
  queue.push(root);
  while (queue.length && root !== null) {
    // 存放当前层级节点数组
    let curLevel = [];
    // 计算当前层级节点数量
    let length = queue.length;
    while (length--) {
      let node = queue.shift();
      // 把当前层节点存入curLevel数组
      curLevel.push(node.val);
      // 把下一层级的左右节点存入queue队列
      node.left && queue.push(node.left);
      node.right && queue.push(node.right);
    }
    // 从数组前头插入值，避免最后反转数组，减少运算时间
    res.unshift(curLevel);
  }
  return res;
};

TypeScript:

function levelOrderBottom(root: TreeNode | null): number[][] {
    let helperQueue: TreeNode[] = [];
    let resArr: number[][] = [];
    let tempArr: number[] = [];
    let tempNode: TreeNode;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            tempNode = helperQueue.shift()!;
            tempArr.push(tempNode.val);
            if (tempNode.left !== null) helperQueue.push(tempNode.left);
            if (tempNode.right !== null) helperQueue.push(tempNode.right);
        }
        resArr.push(tempArr);
        tempArr = [];
    }
    return resArr.reverse();
};

Swift：

func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
    // 表示一层
    var queue = [TreeNode]()
    if let node = root { queue.append(node) }
    var result = [[Int]]()
    while !queue.isEmpty {
        let count = queue.count
        var subarray = [Int]()
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            subarray.append(node.val)
            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node)}
        }
        result.append(subarray)
    }

    return result.reversed()
}

Scala:

// 107.二叉树的层次遍历II
object Solution {
  import scala.collection.mutable
  def levelOrderBottom(root: TreeNode): List[List[Int]] = {
    val res = mutable.ListBuffer[List[Int]]()
    if (root == null) return res.toList
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      val tmp = mutable.ListBuffer[Int]()
      val len = queue.size
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        tmp.append(curNode.value)
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      res.append(tmp.toList)
    }
    // 最后翻转一下
    res.reverse.toList
  }

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn level_order_bottom(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let mut temp = vec![];
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                temp.push(node.borrow().val);
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
            res.push(temp);
        }
        res.into_iter().rev().collect()
    }
}

C#

public IList<IList<int>> LevelOrderBottom(TreeNode root)
{
    var res = new List<IList<int>>();
    var que = new Queue<TreeNode>();
    if (root == null) return res;
    que.Enqueue(root);
    while (que.Count != 0)
    {
        var size = que.Count;
        var vec = new List<int>();
        for (int i = 0; i < size; i++)
        {
            var cur = que.Dequeue();
            vec.Add(cur.val);
            if (cur.left != null) que.Enqueue(cur.left);
            if (cur.right != null) que.Enqueue(cur.right);
        }
        res.Add(vec);
    }
    res.Reverse();
    return res;
}

199.二叉树的右视图

力扣题目链接 (opens new window)

给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

思路

层序遍历的时候，判断是否遍历到单层的最后面的元素，如果是，就放进result数组中，随后返回result就可以了。

C++代码：

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<int> result;
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (i == (size - 1)) result.push_back(node->val); // 将每一层的最后元素放入result数组中
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return result;
    }
};

其他语言版本

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        
        queue = collections.deque([root])
        right_view = []
        
        while queue:
            level_size = len(queue)
            
            for i in range(level_size):
                node = queue.popleft()
                
                if i == level_size - 1:
                    right_view.append(node.val)
                
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        
        return right_view

Java:

// 199.二叉树的右视图
public class N0199 {
    /**
     * 解法：队列，迭代。
     * 每次返回每层的最后一个字段即可。
     *
     * 小优化：每层右孩子先入队。代码略。
     */
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        Deque<TreeNode> que = new LinkedList<>();

        if (root == null) {
            return list;
        }

        que.offerLast(root);
        while (!que.isEmpty()) {
            int levelSize = que.size();

            for (int i = 0; i < levelSize; i++) {
                TreeNode poll = que.pollFirst();

                if (poll.left != null) {
                    que.addLast(poll.left);
                }
                if (poll.right != null) {
                    que.addLast(poll.right);
                }

                if (i == levelSize - 1) {
                    list.add(poll.val);
                }
            }
        }

        return list;
    }
}

Go:

/**
199. 二叉树的右视图
 */
func rightSideView(root *TreeNode) []int {
    if root == nil {
        return nil
    }
    res := make([]int, 0)
    queue := list.New()
    queue.PushBack(root)

    for queue.Len() > 0 {
        length := queue.Len()
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            // 取每层的最后一个元素，添加到结果集中
            if i == length-1 {
                res = append(res, node.Val)
            }
        }
    }
    return res
}

// 使用切片作为队列
func rightSideView(root *TreeNode) []int {
    res := make([]int, 0)
    if root == nil {
        return res
    }
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
            if i == size-1 {
                res = append(res, node.Val)
            }
        }
    }
    return res
}

Javascript:

var rightSideView = function(root) {
    //二叉树右视图 只需要把每一层最后一个节点存储到res数组
    let res = [], queue = [];
    queue.push(root);

    while(queue.length && root!==null) {
        // 记录当前层级节点个数
        let length = queue.length;
        while(length--) {
            let node = queue.shift();
            // length长度为0的时候表明到了层级最后一个节点
            if(!length) {
                res.push(node.val);
            }
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
        }
    }

    return res;
};

TypeScript：

function rightSideView(root: TreeNode | null): number[] {
    let helperQueue: TreeNode[] = [];
    let resArr: number[] = [];
    let tempNode: TreeNode;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            tempNode = helperQueue.shift()!;
            if (i === length - 1) resArr.push(tempNode.val);
            if (tempNode.left !== null) helperQueue.push(tempNode.left);
            if (tempNode.right !== null) helperQueue.push(tempNode.right);
        }
    }
    return resArr;
};

Swift：

func rightSideView(_ root: TreeNode?) -> [Int] {
    // 表示一层
    var queue = [TreeNode]()
    if let node = root { queue.append(node) }
    var result = [Int]()
    while !queue.isEmpty {
        let count = queue.count
        for i in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            if i == count - 1 { result.append(node.val) }

            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
    }

    return result
}

Scala:

// 199.二叉树的右视图
object Solution {
  import scala.collection.mutable
  def rightSideView(root: TreeNode): List[Int] = {
    val res = mutable.ListBuffer[Int]()
    if (root == null) return res.toList
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      val len = queue.size
      var curNode: TreeNode = null
      for (i <- 0 until len) {
        curNode = queue.dequeue()
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      res.append(curNode.value)   // 把最后一个节点的值加入解集
    }
    res.toList  // 最后需要把res转换为List，return关键字可以省略
  }
}

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let len = queue.len();
            for i in 0..len {
                let node = queue.pop_front().unwrap().unwrap();
                if i == len - 1 {
                    res.push(node.borrow().val);
                }
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
        }
        res
    }
}

637.二叉树的层平均值

力扣题目链接 (opens new window)

给定一个非空二叉树, 返回一个由每层节点平均值组成的数组。

思路

本题就是层序遍历的时候把一层求个总和再取一个均值。

C++代码:

class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<double> result;
        while (!que.empty()) {
            int size = que.size();
            double sum = 0; // 统计每一层的和
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                sum += node->val;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(sum / size); // 将每一层均值放进结果集
        }
        return result;
    }
};

其他语言版本

Python：

class Solution:
    """二叉树层平均值迭代解法"""

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfLevels(self, root: TreeNode) -> List[float]:
        if not root:
            return []

        queue = collections.deque([root])
        averages = []
        
        while queue:
            size = len(queue)
            level_sum = 0
            
            for i in range(size):
                node = queue.popleft()
                
                
                level_sum += node.val
                    
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            
            averages.append(level_sum / size)
        
        return averages

Java:

// 637. 二叉树的层平均值
public class N0637 {

    /**
     * 解法：队列，迭代。
     * 每次返回每层的最后一个字段即可。
     */
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> list = new ArrayList<>();
        Deque<TreeNode> que = new LinkedList<>();

        if (root == null) {
            return list;
        }

        que.offerLast(root);
        while (!que.isEmpty()) {

            int levelSize = que.size();
            double levelSum = 0.0;
            for (int i = 0; i < levelSize; i++) {
                TreeNode poll = que.pollFirst();

                levelSum += poll.val;

                if (poll.left != null) {
                    que.addLast(poll.left);
                }
                if (poll.right != null) {
                    que.addLast(poll.right);
                }
            }
            list.add(levelSum / levelSize);
        }
        return list;
    }
}

Go:

/**
637. 二叉树的层平均值
 */
func averageOfLevels(root *TreeNode) []float64 {
    if root == nil {
        // 防止为空
        return nil
    }
    res := make([]float64, 0)
    queue := list.New()
    queue.PushBack(root)

    var sum int
    for queue.Len() > 0 {
        //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数）
        length := queue.Len()
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            // 当前层元素求和
            sum += node.Val
        }
        // 计算每层的平均值，将结果添加到响应结果中
        res = append(res, float64(sum)/float64(length))
        sum = 0 // 清空该层的数据
    }
    return res
}

// 使用切片作为队列
func averageOfLevels(root *TreeNode) []float64 {
    res := make([]float64, 0)
    if root == nil {
        return res
    }
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        sum := 0
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            sum += node.Val
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        res = append(res, float64(sum)/float64(size))
    }
    return res
}

Javascript：

var averageOfLevels = function(root) {
  let res = [],
    queue = [];
  queue.push(root);
  while (queue.length) {
    // 每一层节点个数;
    let lengthLevel = queue.length,
      len = queue.length,
      //   sum记录每一层的和;
      sum = 0;
    while (lengthLevel--) {
      const node = queue.shift();
      sum += node.val;
      //   队列存放下一层节点
      node.left && queue.push(node.left);
      node.right && queue.push(node.right);
    }
    // 求平均值
    res.push(sum / len);
  }
  return res;
};

TypeScript：

function averageOfLevels(root: TreeNode | null): number[] {
    let helperQueue: TreeNode[] = [];
    let resArr: number[] = [];
    let total: number = 0;
    let tempNode: TreeNode;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        let length = helperQueue.length;
        for (let i = 0; i < length; i++) {
            tempNode = helperQueue.shift()!;
            total += tempNode.val;
            if (tempNode.left) helperQueue.push(tempNode.left);
            if (tempNode.right) helperQueue.push(tempNode.right);
        }
        resArr.push(total / length);
        total = 0;
    }
    return resArr;
};

Swift：

func averageOfLevels(_ root: TreeNode?) -> [Double] {
    // 表示一层
    var queue = [TreeNode]()
    if let node = root { queue.append(node) }
    var result = [Double]()
    while !queue.isEmpty {
        let count = queue.count
        var sum = 0
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            sum += node.val

            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
        result.append(Double(sum) / Double(count))
    }

    return result
}

Scala:

// 637.二叉树的层平均值
object Solution {
  import scala.collection.mutable
  def averageOfLevels(root: TreeNode): Array[Double] = {
    val res = mutable.ArrayBuffer[Double]()
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      var sum = 0.0
      var len = queue.size
      for (i <- 0 until len) {
        var curNode = queue.dequeue()
        sum += curNode.value // 累加该层的值
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      res.append(sum / len) // 平均值即为sum/len
    }
    res.toArray // 最后需要转换为Array，return关键字可以省略
  }
}

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn average_of_levels(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<f64> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let len = queue.len();
            let mut sum = 0;
            for _ in 0..len {
                let node = queue.pop_front().unwrap().unwrap();
                sum += node.borrow().val;
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
            res.push((sum as f64) / len as f64);
        }
        res
    }
}

429.N叉树的层序遍历

力扣题目链接 (opens new window)

给定一个 N 叉树，返回其节点值的层序遍历。 (即从左到右，逐层遍历)。

例如，给定一个 3叉树 :

返回其层序遍历:

[ [1], [3,2,4], [5,6] ]

思路

这道题依旧是模板题，只不过一个节点有多个孩子了

C++代码:

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> result;
        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;
            for (int i = 0; i < size; i++) {
                Node* node = que.front();
                que.pop();
                vec.push_back(node->val);
                for (int i = 0; i < node->children.size(); i++) { // 将节点孩子加入队列
                    if (node->children[i]) que.push(node->children[i]);
                }
            }
            result.push_back(vec);
        }
        return result;

    }
};

其他语言版本

Python：

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root:
            return []

        result = []
        queue = collections.deque([root])

        while queue:
            level_size = len(queue)
            level = []

            for _ in range(level_size):
                node = queue.popleft()
                level.append(node.val)

                for child in node.children:
                    queue.append(child)

            result.append(level)

        return result

# LeetCode 429. N-ary Tree Level Order Traversal
# 递归法
class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root: return []
        result=[]
        def traversal(root,depth):
            if len(result)==depth:result.append([])
            result[depth].append(root.val)
            if root.children:
                for i in range(len(root.children)):traversal(root.children[i],depth+1)

        traversal(root,0)
        return result

Java:

// 429. N 叉树的层序遍历
public class N0429 {
    /**
     * 解法1：队列，迭代。
     */
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> list = new ArrayList<>();
        Deque<Node> que = new LinkedList<>();

        if (root == null) {
            return list;
        }

        que.offerLast(root);
        while (!que.isEmpty()) {
            int levelSize = que.size();
            List<Integer> levelList = new ArrayList<>();

            for (int i = 0; i < levelSize; i++) {
                Node poll = que.pollFirst();

                levelList.add(poll.val);

                List<Node> children = poll.children;
                if (children == null || children.size() == 0) {
                    continue;
                }
                for (Node child : children) {
                    if (child != null) {
                        que.offerLast(child);
                    }
                }
            }
            list.add(levelList);
        }

        return list;
    }

    class Node {
        public int val;
        public List<Node> children;

        public Node() {}

        public Node(int _val) {
            val = _val;
        }

        public Node(int _val, List<Node> _children) {
            val = _val;
            children = _children;
        }
    }
}

Go:

/**
429. N 叉树的层序遍历
 */

func levelOrder(root *Node) [][]int {
    queue := list.New()
    res := [][]int{}       //结果集
    if root == nil{
        return res
    }
    queue.PushBack(root)
    for queue.Len() > 0 {
        length := queue.Len()     //记录当前层的数量
        var tmp []int
        for T := 0; T < length; T++ {        //该层的每个元素：一添加到该层的结果集中；二找到该元素的下层元素加入到队列中，方便下次使用
            myNode := queue.Remove(queue.Front()).(*Node)
            tmp = append(tmp, myNode.Val)
            for i := 0; i < len(myNode.Children); i++ {
                queue.PushBack(myNode.Children[i])
            }
        }
        res = append(res, tmp)
    }

    return res
}

// 使用切片作为队列
func levelOrder(root *Node) [][]int {
    res := make([][]int, 0)
    if root == nil {
        return res
    }
    queue := make([]*Node, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        level := make([]int, 0)
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            level = append(level, node.Val)
            if len(node.Children) > 0 {
                queue = append(queue, node.Children...)
            }
        }
        res = append(res, level)
    }
    return res
}

JavaScript：

var levelOrder = function(root) {
    //每一层可能有2个以上,所以不再使用node.left node.right
    let res = [], queue = [];
    queue.push(root);

    while(queue.length && root!==null) {
        //记录每一层节点个数还是和二叉树一致
        let length = queue.length;
        //存放每层节点 也和二叉树一致
        let curLevel = [];
        while(length--) {
            let node = queue.shift();
            curLevel.push(node.val);

            //这里不再是 ndoe.left node.right 而是循坏node.children
           for(let item of node.children){
               item && queue.push(item);
           }
        }
        res.push(curLevel);
    }

    return res;
};

TypeScript:

function levelOrder(root: Node | null): number[][] {
    let helperQueue: Node[] = [];
    let resArr: number[][] = [];
    let tempArr: number[] = [];
    if (root !== null) helperQueue.push(root);
    let curNode: Node;
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            curNode = helperQueue.shift()!;
            tempArr.push(curNode.val);
            helperQueue.push(...curNode.children);
        }
        resArr.push(tempArr);
        tempArr = [];
    }
    return resArr;
};

Swift：

func levelOrder(_ root: Node?) -> [[Int]] {
    // 表示一层
    var queue = [Node]()
    if let node = root { queue.append(node) }
    var result = [[Int]]()
    while !queue.isEmpty {
        let count = queue.count
        var subarray = [Int]()
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            subarray.append(node.val)
            // 下一层
            for node in node.children { queue.append(node) }
        }
        result.append(subarray)
    }

    return result
}

Scala:

// 429.N叉树的层序遍历
object Solution {
  import scala.collection.mutable
  def levelOrder(root: Node): List[List[Int]] = {
    val res = mutable.ListBuffer[List[Int]]()
    if (root == null) return res.toList
    val queue = mutable.Queue[Node]()
    queue.enqueue(root) // 根节点入队
    while (!queue.isEmpty) {
      val tmp = mutable.ListBuffer[Int]() // 存储每层节点
      val len = queue.size
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        tmp.append(curNode.value) // 将该节点的值加入tmp
        // 循环遍历该节点的子节点，加入队列
        for (child <- curNode.children) {
          queue.enqueue(child)
        }
      }
      res.append(tmp.toList) // 将该层的节点放到结果集
    }
    res.toList
  }
}

Rust:

pub struct Solution;
#[derive(Debug, PartialEq, Eq)]
pub struct Node {
    pub val: i32,
    pub children: Vec<Option<Rc<RefCell<Node>>>>,
}

impl Node {
    #[inline]
    pub fn new(val: i32) -> Node {
        Node {
            val,
            children: vec![],
        }
    }
}

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn level_order(root: Option<Rc<RefCell<Node>>>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let mut temp = vec![];
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                temp.push(node.borrow().val);
                if !node.borrow().children.is_empty() {
                    for n in node.borrow().children.clone() {
                        queue.push_back(n);
                    }
                }
            }
            res.push(temp)
        }
        res
    }
}

515.在每个树行中找最大值

力扣题目链接 (opens new window)

您需要在二叉树的每一行中找到最大的值。

思路

层序遍历，取每一层的最大值

C++代码：

class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<int> result;
        while (!que.empty()) {
            int size = que.size();
            int maxValue = INT_MIN; // 取每一层的最大值
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                maxValue = node->val > maxValue ? node->val : maxValue;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(maxValue); // 把最大值放进数组
        }
        return result;
    }
};

其他语言版本

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def largestValues(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        result = []
        queue = collections.deque([root])

        while queue:
            level_size = len(queue)
            max_val = float('-inf')

            for _ in range(level_size):
                node = queue.popleft()
                max_val = max(max_val, node.val)

                if node.left:
                    queue.append(node.left)

                if node.right:
                    queue.append(node.right)

            result.append(max_val)

        return result

Java：

class Solution {
    public List<Integer> largestValues(TreeNode root) {
        if(root == null){
            return Collections.emptyList();
        }
        List<Integer> result = new ArrayList();
        Queue<TreeNode> queue = new LinkedList();
        queue.offer(root);
        while(!queue.isEmpty()){
            int max = Integer.MIN_VALUE;
            for(int i = queue.size(); i > 0; i--){
               TreeNode node = queue.poll();
               max = Math.max(max, node.val);
               if(node.left != null) queue.offer(node.left);
               if(node.right != null) queue.offer(node.right);
            }
            result.add(max);
        }
        return result;
    }
}

Go:

/**
515. 在每个树行中找最大值
 */
func largestValues(root *TreeNode) []int {
    if root == nil {
        //防止为空
        return nil
    }
    queue := list.New()
    queue.PushBack(root)
    ans := make([]int, 0)
    temp := math.MinInt64
    // 层序遍历
    for queue.Len() > 0 {
        //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数）
        length := queue.Len()
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)//出队列
            // 比较当前层中的最大值和新遍历的元素大小，取两者中大值
            temp = max(temp, node.Val)
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
        }
        ans = append(ans, temp)
        temp = math.MinInt64
    }
    return ans
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

// 使用切片作为队列
func largestValues(root *TreeNode) []int {
    res := make([]int, 0)
    if root == nil {
        return res
    }
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        maxValue := math.MinInt64
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Val > maxValue {
                maxValue = node.Val
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        res = append(res, maxValue)
    }
    return res
}

Javascript：

var largestValues = function (root) {
  let res = [],
    queue = [];
  queue.push(root);
  if (root === null) {
    return res;
  }
  while (queue.length) {
    let lengthLevel = queue.length,
      // 初始值设为负无穷大
      max = -Infinity;
    while (lengthLevel--) {
      const node = queue.shift();
      //   在当前层中找到最大值
      max = Math.max(max, node.val);
      // 找到下一层的节点
      node.left && queue.push(node.left);
      node.right && queue.push(node.right);
    }
    res.push(max);
  }
  return res;
};

TypeScript:

function largestValues(root: TreeNode | null): number[] {
    let helperQueue: TreeNode[] = [];
    let resArr: number[] = [];
    let tempNode: TreeNode;
    let max: number = 0;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            tempNode = helperQueue.shift()!;
            if (i === 0) {
                max = tempNode.val;
            } else {
                max = max > tempNode.val ? max : tempNode.val;
            }
            if (tempNode.left) helperQueue.push(tempNode.left);
            if (tempNode.right) helperQueue.push(tempNode.right);
        }
        resArr.push(max);
    }
    return resArr;
};

Swift：

func largestValues(_ root: TreeNode?) -> [Int] {
    // 表示一层
    var queue = [TreeNode]()
    if let node = root { queue.append(node) }
    var result = [Int]()
    while !queue.isEmpty {
        let count = queue.count
        var max = queue[0].val
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            if node.val > max { max = node.val }

            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
        result.append(max)
    }

    return result
}

Scala:

// 515.在每个树行中找最大值
object Solution {
  import scala.collection.mutable
  def largestValues(root: TreeNode): List[Int] = {
    val res = mutable.ListBuffer[Int]()
    if (root == null) return res.toList
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      var max = Int.MinValue // 初始化max为系统最小值
      val len = queue.size
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        max = math.max(max, curNode.value) // 对比求解最大值
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      res.append(max) // 将最大值放入结果集
    }
    res.toList
  }
}

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = vec![];
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            let mut max = i32::MIN;
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                max = max.max(node.borrow().val);
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
            res.push(max);
        }
        res
    }
}

116.填充每个节点的下一个右侧节点指针

力扣题目链接 (opens new window)

给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

思路

本题依然是层序遍历，只不过在单层遍历的时候记录一下本层的头部节点，然后在遍历的时候让前一个节点指向本节点就可以了

C++代码：

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            // vector<int> vec;
            Node* nodePre;
            Node* node;
            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front(); // 取出一层的头结点
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node; // 本层前一个节点next指向本节点
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL; // 本层最后一个节点指向NULL
        }
        return root;

    }
};

其他语言版本

Java：

class Solution {
    public Node connect(Node root) {
	Queue<Node> tmpQueue = new LinkedList<Node>();
	if (root != null) tmpQueue.add(root);

	while (tmpQueue.size() != 0){
	    int size = tmpQueue.size();

            Node cur = tmpQueue.poll();
            if (cur.left != null) tmpQueue.add(cur.left);
            if (cur.right != null) tmpQueue.add(cur.right);

	    for (int index = 1; index < size; index++){
		Node next = tmpQueue.poll();
		if (next.left != null) tmpQueue.add(next.left);
		if (next.right != null) tmpQueue.add(next.right);

                cur.next = next;
                cur = next;
	    }
	}

        return root;
    }
}

Python：

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return root
        
        queue = collections.deque([root])
        
        while queue:
            level_size = len(queue)
            prev = None
            
            for i in range(level_size):
                node = queue.popleft()
                
                if prev:
                    prev.next = node
                
                prev = node
                
                if node.left:
                    queue.append(node.left)
                
                if node.right:
                    queue.append(node.right)
        
        return root

Go:

/**
116. 填充每个节点的下一个右侧节点指针
117. 填充每个节点的下一个右侧节点指针 II
 */

func connect(root *Node) *Node {
    if root == nil { //防止为空
        return root
    }
    queue := list.New()
    queue.PushBack(root)
    tmpArr := make([]*Node, 0)
    for queue.Len() > 0 {
        length := queue.Len() //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数）
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*Node) //出队列
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            tmpArr = append(tmpArr, node) //将值加入本层切片中
        }
        if len(tmpArr) > 1 {
            // 遍历每层元素,指定next
            for i := 0; i < len(tmpArr)-1; i++ {
                tmpArr[i].Next = tmpArr[i+1]
            }
        }
        tmpArr = []*Node{} //清空层的数据
    }
    return root
}

// 使用切片作为队列
func connect(root *Node) *Node {
    if root == nil {
        return root
    }
    queue := make([]*Node, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            node := queue[i]
            if i != size - 1 {
                queue[i].Next = queue[i+1]
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        queue = queue[size:]
    }
    return root
}

JavaScript:

/**
 * // Definition for a Node.
 * function Node(val, left, right, next) {
 *    this.val = val === undefined ? null : val;
 *    this.left = left === undefined ? null : left;
 *    this.right = right === undefined ? null : right;
 *    this.next = next === undefined ? null : next;
 * };
 */

/**
 * @param {Node} root
 * @return {Node}
 */
var connect = function(root) {
    if (root === null) return root;
    let queue = [root];
    while (queue.length) {
        let n = queue.length;
        for (let i = 0; i < n; i++) {
            let node = queue.shift();
            if (i < n-1) {
                node.next = queue[0];
            }
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
        }
    }
    return root;
};

TypeScript:

function connect(root: Node | null): Node | null {
    let helperQueue: Node[] = [];
    let preNode: Node, curNode: Node;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            if (i === 0) {
                preNode = helperQueue.shift()!;
            } else {
                curNode = helperQueue.shift()!;
                preNode.next = curNode;
                preNode = curNode;
            }
            if (preNode.left) helperQueue.push(preNode.left);
            if (preNode.right) helperQueue.push(preNode.right);
        }
        preNode.next = null;
    }
    return root;
};

Swift：

func connect(_ root: Node?) -> Node? {
    // 表示一层
    var queue = [Node]()
    if let node = root { queue.append(node) }
    while !queue.isEmpty {
        let count = queue.count
        var current, previous: Node!
        for i in 0 ..< count {
            // 当前层
            if i == 0 {
                previous = queue.removeFirst()
                current = previous
            } else {
                current = queue.removeFirst()
                previous.next = current
                previous = current
            }

            // 下一层
            if let node = current.left { queue.append(node) }
            if let node = current.right { queue.append(node) }
        }
        previous.next = nil
    }

    return root
}

Scala:

// 116.填充每个节点的下一个右侧节点指针
object Solution {
  import scala.collection.mutable

  def connect(root: Node): Node = {
    if (root == null) return root
    val queue = mutable.Queue[Node]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      val len = queue.size
      val tmp = mutable.ListBuffer[Node]()
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        tmp.append(curNode)
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      // 处理next指针
      for (i <- 0 until tmp.size - 1) {
        tmp(i).next = tmp(i + 1)
      }
      tmp(tmp.size-1).next = null
    }
    root
  }
}

117.填充每个节点的下一个右侧节点指针II

力扣题目链接 (opens new window)

思路

这道题目说是二叉树，但116题目说是完整二叉树，其实没有任何差别，一样的代码一样的逻辑一样的味道

C++代码：

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;
            Node* nodePre;
            Node* node;
            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front(); // 取出一层的头结点
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node; // 本层前一个节点next指向本节点
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL; // 本层最后一个节点指向NULL
        }
        return root;
    }
};

其他语言版本

Java：

// 二叉树之层次遍历
class Solution {
    public Node connect(Node root) {
        Queue<Node> queue = new LinkedList<>();
        if (root != null) {
            queue.add(root);
        }
        while (!queue.isEmpty()) {
            int size = queue.size();
            Node node = null;
            Node nodePre = null;

            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = queue.poll(); // 取出本层头一个节点
                    node = nodePre;
                } else {
                    node = queue.poll();
                    nodePre.next = node; // 本层前一个节点 next 指向当前节点
                    nodePre = nodePre.next;
                }
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            nodePre.next = null; // 本层最后一个节点 next 指向 null
        }
        return root;
    }
}

Python：

# 层序遍历解法
"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return root
        
        queue = collections.deque([root])
        
        while queue:
            level_size = len(queue)
            prev = None
            
            for i in range(level_size):
                node = queue.popleft()
                
                if prev:
                    prev.next = node
                
                prev = node
                
                if node.left:
                    queue.append(node.left)
                
                if node.right:
                    queue.append(node.right)
        
        return root

Go:

/**
116. 填充每个节点的下一个右侧节点指针
117. 填充每个节点的下一个右侧节点指针 II
 */

func connect(root *Node) *Node {
    if root == nil { //防止为空
        return root
    }
    queue := list.New()
    queue.PushBack(root)
    tmpArr := make([]*Node, 0)
    for queue.Len() > 0 {
        length := queue.Len() //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数）
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*Node) //出队列
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
            tmpArr = append(tmpArr, node) //将值加入本层切片中
        }
        if len(tmpArr) > 1 {
            // 遍历每层元素,指定next
            for i := 0; i < len(tmpArr)-1; i++ {
                tmpArr[i].Next = tmpArr[i+1]
            }
        }
        tmpArr = []*Node{} //清空层的数据
    }
    return root
}

// 使用切片作为队列
func connect(root *Node) *Node {
    if root == nil {
        return root
    }
    queue := make([]*Node, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            node := queue[i]
            if i != size - 1 {
                queue[i].Next = queue[i+1]
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        queue = queue[size:]
    }
    return root
}

JavaScript:

/**
 * // Definition for a Node.
 * function Node(val, left, right, next) {
 *    this.val = val === undefined ? null : val;
 *    this.left = left === undefined ? null : left;
 *    this.right = right === undefined ? null : right;
 *    this.next = next === undefined ? null : next;
 * };
 */

/**
 * @param {Node} root
 * @return {Node}
 */
var connect = function(root) {
    if (root === null) {
        return null;
    }
    let queue = [root];
    while (queue.length > 0) {
        let n = queue.length;
        for (let i=0; i<n; i++) {
            let node = queue.shift();
            if (i < n-1) node.next = queue[0];
            if (node.left != null) queue.push(node.left);
            if (node.right != null) queue.push(node.right);
        }
    }
    return root;
};

TypeScript:

function connect(root: Node | null): Node | null {
    let helperQueue: Node[] = [];
    let preNode: Node, curNode: Node;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            if (i === 0) {
                preNode = helperQueue.shift()!;
            } else {
                curNode = helperQueue.shift()!;
                preNode.next = curNode;
                preNode = curNode;
            }
            if (preNode.left) helperQueue.push(preNode.left);
            if (preNode.right) helperQueue.push(preNode.right);
        }
        preNode.next = null;
    }
    return root;
};

Swift：

func connect(_ root: Node?) -> Node? {
    // 表示一层
    var queue = [Node]()
    if let node = root { queue.append(node) }
    while !queue.isEmpty {
        let count = queue.count
        var current, previous: Node!
        for i in 0 ..< count {
            // 当前层
            if i == 0 {
                previous = queue.removeFirst()
                current = previous
            } else {
                current = queue.removeFirst()
                previous.next = current
                previous = current
            }

            // 下一层
            if let node = current.left { queue.append(node) }
            if let node = current.right { queue.append(node) }
        }
        previous.next = nil
    }

    return root
}

Scala:

// 117.填充每个节点的下一个右侧节点指针II
object Solution {
  import scala.collection.mutable

  def connect(root: Node): Node = {
    if (root == null) return root
    val queue = mutable.Queue[Node]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      val len = queue.size
      val tmp = mutable.ListBuffer[Node]()
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        tmp.append(curNode)
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
      // 处理next指针
      for (i <- 0 until tmp.size - 1) {
        tmp(i).next = tmp(i + 1)
      }
      tmp(tmp.size-1).next = null
    }
    root
  }
}

104.二叉树的最大深度

力扣题目链接 (opens new window)

给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例：

给定二叉树 [3,9,20,null,null,15,7]，

返回它的最大深度 3 。

思路

使用迭代法的话，使用层序遍历是最为合适的，因为最大的深度就是二叉树的层数，和层序遍历的方式极其吻合。

在二叉树中，一层一层的来遍历二叉树，记录一下遍历的层数就是二叉树的深度，如图所示：

所以这道题的迭代法就是一道模板题，可以使用二叉树层序遍历的模板来解决的。

C++代码如下：

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty()) {
            int size = que.size();
            depth++; // 记录深度
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return depth;
    }
};

其他语言版本

Java：

class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null)   return 0;
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root);
        int depth = 0;
        while (!que.isEmpty())
        {
            int len = que.size();
            while (len > 0)
            {
                TreeNode node = que.poll();
                if (node.left != null)  que.offer(node.left);
                if (node.right != null) que.offer(node.right);
                len--;
            }
            depth++;
        }
        return depth;
    }
}

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        
        depth = 0
        queue = collections.deque([root])
        
        while queue:
            depth += 1
            for _ in range(len(queue)):
                node = queue.popleft()
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        
        return depth

Go：

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxDepth(root *TreeNode) int {
    ans := 0
    if root == nil {
        return 0
    }
    queue := list.New()
    queue.PushBack(root)
    for queue.Len() > 0 {
        length := queue.Len()
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
        }
        ans++//记录深度，其他的是层序遍历的板子
    }
    return ans
}

// 使用切片作为队列
func maxDepth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    depth := 0
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        depth++
    }
    return depth
}

JavaScript：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxDepth = function (root) {
  // 二叉树的 最大深度 是指从根节点到最远叶子节点的最长路径上的节点数。
  let max = 0,
    queue = [root];
  if (root === null) {
    return max;
  }
  while (queue.length) {
    max++;
    let length = queue.length;
    while (length--) {
      let node = queue.shift();
      node.left && queue.push(node.left);
      node.right && queue.push(node.right);
    }
  }
  return max;
};

TypeScript:

function maxDepth(root: TreeNode | null): number {
    let helperQueue: TreeNode[] = [];
    let resDepth: number = 0;
    let tempNode: TreeNode;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        resDepth++;
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            tempNode = helperQueue.shift()!;
            if (tempNode.left) helperQueue.push(tempNode.left);
            if (tempNode.right) helperQueue.push(tempNode.right);
        }
    }
    return resDepth;
};

Swift：

func maxDepth(_ root: TreeNode?) -> Int {
    guard root != nil else { return 0 }
    var depth = 0
    var queue = [TreeNode]()
    queue.append(root!)
    while !queue.isEmpty {
        let count = queue.count
        depth += 1
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()

            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
    }

    return depth
}

Scala:

// 104.二叉树的最大深度
object Solution {
  import scala.collection.mutable
  def maxDepth(root: TreeNode): Int = {
    if (root == null) return 0
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    var depth = 0
    while (!queue.isEmpty) {
      val len = queue.length
      depth += 1
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
      }
    }
    depth
  }
}

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut queue = VecDeque::new();
        let mut res = 0;
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            res += 1;
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
        }
        res
    }
}

111.二叉树的最小深度

力扣题目链接 (opens new window)

思路

相对于 104.二叉树的最大深度 ，本题还也可以使用层序遍历的方式来解决，思路是一样的。

需要注意的是，只有当左右孩子都为空的时候，才说明遍历的最低点了。如果其中一个孩子为空则不是最低点

代码如下：（详细注释）

class Solution {
public:
    int minDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty()) {
            int size = que.size();
            depth++; // 记录最小深度
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
                if (!node->left && !node->right) { // 当左右孩子都为空的时候，说明是最低点的一层了，退出
                    return depth;
                }
            }
        }
        return depth;
    }
};

其他语言版本

Java：

class Solution {
    public int minDepth(TreeNode root){
        if (root == null) {
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int depth = 0;
        while (!queue.isEmpty()){
            int size = queue.size();
            depth++;
            TreeNode cur = null;
            for (int i = 0; i < size; i++) {
                cur = queue.poll();
                //如果当前节点的左右孩子都为空，直接返回最小深度
                if (cur.left == null && cur.right == null){
                    return depth;
                }
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
        }
        return depth;
    }
}

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        depth = 0
        queue = collections.deque([root])
        
        while queue:
            depth += 1 
            for _ in range(len(queue)):
                node = queue.popleft()
                
                if not node.left and not node.right:
                    return depth
            
                if node.left:
                    queue.append(node.left)
                    
                if node.right:
                    queue.append(node.right)

        return depth

Go：

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDepth(root *TreeNode) int {
    ans := 0
    if root == nil {
        return 0
    }
    queue := list.New()
    queue.PushBack(root)
    for queue.Len() > 0 {
        length := queue.Len()
        for i := 0; i < length; i++ {
            node := queue.Remove(queue.Front()).(*TreeNode)
            if node.Left == nil && node.Right == nil {      //当前节点没有左右节点，则代表此层是最小层
                return ans+1                                //返回当前层 ans代表是上一层
            }
            if node.Left != nil {
                queue.PushBack(node.Left)
            }
            if node.Right != nil {
                queue.PushBack(node.Right)
            }
        }
        ans++//记录层数
    }

    return ans
}

// 使用切片作为队列
func minDepth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    depth := 0
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        depth++
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
            if node.Left == nil && node.Right == nil {
                return depth
            }
        }
    }
    return depth
}

JavaScript：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function(root) {
    if (root === null) return 0;
    let queue = [root];
    let depth = 0;
    while (queue.length) {
        let n = queue.length;
        depth++;
        for (let i=0; i<n; i++) {
            let node = queue.shift();
            // 如果左右节点都是null(在遇见的第一个leaf节点上)，则该节点深度最小
            if (node.left === null && node.right === null) {
                return depth;
            }
            node.left && queue.push(node.left);;
            node.right && queue.push(node.right);
        }
    }
    return depth;
};

TypeScript:

function minDepth(root: TreeNode | null): number {
    let helperQueue: TreeNode[] = [];
    let resMin: number = 0;
    let tempNode: TreeNode;
    if (root !== null) helperQueue.push(root);
    while (helperQueue.length > 0) {
        resMin++;
        for (let i = 0, length = helperQueue.length; i < length; i++) {
            tempNode = helperQueue.shift()!;
            if (tempNode.left === null && tempNode.right === null) return resMin;
            if (tempNode.left !== null) helperQueue.push(tempNode.left);
            if (tempNode.right !== null) helperQueue.push(tempNode.right);
        }
    }
    return resMin;
};

Swift：

func minDepth(_ root: TreeNode?) -> Int {
    guard root != nil else { return 0 }
    var depth = 0
    var queue = [root!]
    while !queue.isEmpty {
        let count = queue.count
        depth += 1
        for _ in 0 ..< count {
            // 当前层
            let node = queue.removeFirst()
            if node.left == nil, node.right == nil { // 遇到叶子结点则返回
                return depth
            }

            // 下一层
            if let node = node.left { queue.append(node) }
            if let node = node.right { queue.append(node) }
        }
    }
    return depth
}

Scala:

// 111.二叉树的最小深度
object Solution {
  import scala.collection.mutable
  def minDepth(root: TreeNode): Int = {
    if (root == null) return 0
    var depth = 0
    val queue = mutable.Queue[TreeNode]()
    queue.enqueue(root)
    while (!queue.isEmpty) {
      depth += 1
      val len = queue.size
      for (i <- 0 until len) {
        val curNode = queue.dequeue()
        if (curNode.left != null) queue.enqueue(curNode.left)
        if (curNode.right != null) queue.enqueue(curNode.right)
        if (curNode.left == null && curNode.right == null) return depth
      }
    }
    depth
  }
}

Rust:

use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut res = 0;
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root);
        }
        while !queue.is_empty() {
            res += 1;
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap().unwrap();
                if node.borrow().left.is_none() && node.borrow().right.is_none() {
                    return res;
                }
                if node.borrow().left.is_some() {
                    queue.push_back(node.borrow().left.clone());
                }
                if node.borrow().right.is_some() {
                    queue.push_back(node.borrow().right.clone());
                }
            }
        }
        res
    }
}

总结

二叉树的层序遍历，就是图论中的广度优先搜索在二叉树中的应用，需要借助队列来实现（此时又发现队列的一个应用了）。

来吧，一口气打十个：

• 102.二叉树的层序遍历 (opens new window)
• 107.二叉树的层次遍历II (opens new window)
• 199.二叉树的右视图 (opens new window)
• 637.二叉树的层平均值 (opens new window)
• 429.N叉树的层序遍历 (opens new window)
• 515.在每个树行中找最大值 (opens new window)
• 116.填充每个节点的下一个右侧节点指针 (opens new window)
• 117.填充每个节点的下一个右侧节点指针II (opens new window)
• 104.二叉树的最大深度 (opens new window)
• 111.二叉树的最小深度 (opens new window)

致敬叶师傅！