110. Balanced Binary Tree（判断是否为平衡二叉树）_he function isbalanced is to judge if a given bina-优快云博客

本文介绍了一种高效判断二叉树是否平衡的算法。平衡二叉树定义为任意节点的左右子树深度差不超过1的二叉树。文章提供了两种方法：第一种使用两次递归分别计算树的深度和平衡性；第二种为优化版，仅用一次递归同时判断深度与平衡，提高了运行效率。

题目描述

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

在这里插入图片描述

方法思路

Approach1:两个递归

class Solution {
    //Runtime: 1 ms, faster than 86.76%
    //Memory Usage: 37.6 MB, less than 94.33%
    public boolean isBalanced(TreeNode root) {
        if(root == null || (root.left == null && root.right == null))
            return true;
        boolean flag = true;
        int left_depth = 0, right_depth = 0;
        left_depth = maxDepth(root.left);
        right_depth = maxDepth(root.right);
        if(Math.abs(left_depth - right_depth) > 1)
            flag = false;
        else
            flag = true;
        return flag && isBalanced(root.left) && isBalanced(root.right);
    }
    
    public int maxDepth(TreeNode root){
        if(root == null) return 0;
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}

Approach2:方法一的优化版本，只有一个递归

public class Solution {
    //Runtime: 0 ms, faster than 100.00%
    //Memory Usage: 40.7 MB, less than 26.71%
    private boolean result = true;

    public boolean isBalanced(TreeNode root) {
        maxDepth(root);
        return result;
    }

    public int maxDepth(TreeNode root) {
        if (root == null)
            return 0;
        int l = maxDepth(root.left);
        int r = maxDepth(root.right);
        if (Math.abs(l - r) > 1)
            result = false;
        return 1 + Math.max(l, r);
    }
}