参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!

# 1382.将二叉搜索树变平衡

力扣题目链接 (opens new window)

给你一棵二叉搜索树,请你返回一棵 平衡后 的二叉搜索树,新生成的树应该与原来的树有着相同的节点值。

如果一棵二叉搜索树中,每个节点的两棵子树高度差不超过 1 ,我们就称这棵二叉搜索树是 平衡的 。

如果有多种构造方法,请你返回任意一种。

示例:

  • 输入:root = [1,null,2,null,3,null,4,null,null]
  • 输出:[2,1,3,null,null,null,4]
  • 解释:这不是唯一的正确答案,[3,1,4,null,2,null,null] 也是一个可行的构造方案。

提示:

  • 树节点的数目在 1 到 10^4 之间。
  • 树节点的值互不相同,且在 1 到 10^5 之间。

# 思路

这道题目,可以中序遍历把二叉树转变为有序数组,然后在根据有序数组构造平衡二叉搜索树。

建议做这道题之前,先看如下两篇题解:

这两道题目做过之后,本题分分钟就可以做出来了。

代码如下:

class Solution {
private:
    vector<int> vec;
    // 有序树转成有序数组
    void traversal(TreeNode* cur) {
        if (cur == nullptr) {
            return;
        }
        traversal(cur->left);
        vec.push_back(cur->val);
        traversal(cur->right);
    }
    // 有序数组转平衡二叉树
    TreeNode* getTree(vector<int>& nums, int left, int right) {
        if (left > right) return nullptr;
        int mid = left + ((right - left) / 2);
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = getTree(nums, left, mid - 1);
        root->right = getTree(nums, mid + 1, right);
        return root;
    }

public:
    TreeNode* balanceBST(TreeNode* root) {
        traversal(root);
        return getTree(vec, 0, vec.size() - 1);
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

# 其他语言版本

# Java:

class Solution {
    ArrayList <Integer> res = new ArrayList<Integer>();
    // 有序树转成有序数组
    private void travesal(TreeNode cur) {
            if (cur == null) return;
            travesal(cur.left);
            res.add(cur.val);
            travesal(cur.right);
        }
    // 有序数组转成平衡二叉树
    private TreeNode getTree(ArrayList <Integer> nums, int left, int right) {
        if (left > right) return null;
        int mid = left + (right - left) / 2;
        TreeNode root = new TreeNode(nums.get(mid));
        root.left = getTree(nums, left, mid - 1);
        root.right = getTree(nums, mid + 1, right);
        return root;
    }
    public TreeNode balanceBST(TreeNode root) {
        travesal(root);
        return getTree(res, 0, res.size() - 1);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

# Python:

class Solution:
    def balanceBST(self, root: TreeNode) -> TreeNode:
        res = []
        # 有序树转成有序数组
        def traversal(cur: TreeNode):
            if not cur: return
            traversal(cur.left)
            res.append(cur.val)
            traversal(cur.right)
        # 有序数组转成平衡二叉树
        def getTree(nums: List, left, right):
            if left > right: return 
            mid = left + (right -left) // 2
            root = TreeNode(nums[mid])
            root.left = getTree(nums, left, mid - 1)
            root.right = getTree(nums, mid + 1, right)
            return root
        traversal(root)
        return getTree(res, 0, len(res) - 1)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

# Go:

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func balanceBST(root *TreeNode) *TreeNode {
   // 二叉搜索树中序遍历得到有序数组
	nums := []int{}
   // 中序递归遍历二叉树
	var travel func(node *TreeNode)
	travel = func(node *TreeNode) {
		if node == nil {
			return
		}
		travel(node.Left)
		nums = append(nums, node.Val)
		travel(node.Right)
	}
	// 二分法保证左右子树高度差不超过一(题目要求返回的仍是二叉搜索树)
	var buildTree func(nums []int, left, right int) *TreeNode
	buildTree = func(nums []int, left, right int) *TreeNode {
		if left > right {
			return nil
		}
		mid := left + (right-left) >> 1
		root := &TreeNode{Val: nums[mid]}
		root.Left = buildTree(nums, left, mid-1)
		root.Right = buildTree(nums, mid+1, right)
		return root
	}
	travel(root)
	return buildTree(nums, 0, len(nums)-1)
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

# JavaScript:

var balanceBST = function(root) {
    const res = [];
    // 中序遍历转成有序数组
    const travesal = cur => {
        if(!cur) return;
        travesal(cur.left);
        res.push(cur.val);
        travesal(cur.right);
    }
    // 有序数组转成平衡二叉树
    const getTree = (nums, left, right) => {
        if(left > right) return null;
        let mid = left + ((right - left) >> 1);
        let root = new TreeNode(nums[mid]);// 中心位置作为当前节点的值
        root.left = getTree(nums, left, mid - 1);// 递归地将区间[left,mid−1] 作为当前节点的左子树
        root.right = getTree(nums, mid + 1, right);// 递归地将区间[mid+1,right] 作为当前节点的左子树
        return root;
    }
    travesal(root);
    return getTree(res, 0, res.length - 1);
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

# TypeScript:

function balanceBST(root: TreeNode | null): TreeNode | null {
    const inorderArr: number[] = [];
    inorderTraverse(root, inorderArr);
    return buildTree(inorderArr, 0, inorderArr.length - 1);
};
function inorderTraverse(node: TreeNode | null, arr: number[]): void {
    if (node === null) return;
    inorderTraverse(node.left, arr);
    arr.push(node.val);
    inorderTraverse(node.right, arr);
}
function buildTree(arr: number[], left: number, right: number): TreeNode | null {
    if (left > right) return null;
    const mid = (left + right) >> 1;
    const resNode: TreeNode = new TreeNode(arr[mid]);
    resNode.left = buildTree(arr, left, mid - 1);
    resNode.right = buildTree(arr, mid + 1, right);
    return resNode;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

上次更新:: 9/19/2023, 4:22:40 PM
@2021-2022 代码随想录 版权所有 粤ICP备19156078号