【LeetCode】110. Balanced Binary Tree(Easy)解题报告_leetcode balanced binary tree c++-优快云博客

本文链接：https://blog.youkuaiyun.com/haoxiaoxiaoyu/article/details/79595165

本文针对LeetCode上的平衡二叉树问题提供了解决方案，采用后序遍历的方式检查每个节点的左右子树高度差，确保其不超过1，从而判断整棵树是否为高度平衡的二叉树。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

【LeetCode】110. Balanced Binary Tree(Easy)解题报告

题目地址：https://leetcode.com/problems/balanced-binary-tree/description/
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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7
Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4
Return false.

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 判断是不是高度平衡，左右子树高度差值是否大于一
 后序遍历判断左右子树高度
 time : O(n)
 space : O(n)
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) return true;
        return helper(root)!=-1;
    }
    public int helper(TreeNode root){
        if(root == null) return 0;
        int l = helper(root.left);
        int r = helper(root.right);
        if(l == -1 || r == -1 || Math.abs(l-r)>1){
            return -1;
        }
        return Math.max(l,r)+1;
    }
}