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110.平衡二叉树.js
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110.平衡二叉树.js
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/*
* @lc app=leetcode.cn id=110 lang=javascript
*
* [110] 平衡二叉树
*
* https://leetcode-cn.com/problems/balanced-binary-tree/description/
*
* algorithms
* Easy (56.30%)
* Likes: 776
* Dislikes: 0
* Total Accepted: 256.7K
* Total Submissions: 455.7K
* Testcase Example: '[3,9,20,null,null,15,7]'
*
* 给定一个二叉树,判断它是否是高度平衡的二叉树。
*
* 本题中,一棵高度平衡二叉树定义为:
*
*
* 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。
*
*
*
*
* 示例 1:
*
*
* 输入:root = [3,9,20,null,null,15,7]
* 输出:true
*
*
* 示例 2:
*
*
* 输入:root = [1,2,2,3,3,null,null,4,4]
* 输出:false
*
*
* 示例 3:
*
*
* 输入:root = []
* 输出:true
*
*
*
*
* 提示:
*
*
* 树中的节点数在范围 [0, 5000] 内
* -10^4
*
*
*/
// @lc code=start
/**
* 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 {boolean}
*/
var isBalanced = function(root) {
const dfs=(node,prevH=0)=>{
if(node){
const {h:lh,balance:lb}=dfs(node.left,prevH+1)
const {h:rh,balance:rb}=dfs(node.right,prevH+1)
const balance=lb&&rb&&Math.abs(lh-rh)<=1
const h=Math.max(lh,rh)
return {h,balance}
}
return {h:prevH, balance:true}
}
return dfs(root).balance
};
// @lc code=end