Leetcode 110. Balanced Binary Tree 平衡二叉树 解题报告

1 解题思想

题目意思是给定一颗树，判断是否高度平衡，即左右子树的高度差不超过1

采用先序的方式递归遍历到最底层，从最底层开始检查高度是否满足条件，左右的高度是否差值超过1，要是超过了就直接return了。

2 原题

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.

3 AC解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 /**
  * 递归判断每一颗子树是否满足，并把自己的高度返回
  * 每棵树平衡的定义是，左子树的高度和又子数的高度相差不超过1
  * 
  * 只要任何一个地方不满足了，就要break
  * */
public class Solution {
    boolean isBalanced = true;
    public int dfs(TreeNode root){
        if(isBalanced==false)
            return -1;
        if(root==null) return 0;
        int left=dfs(root.left);
        int right=dfs(root.right);
        if(Math.abs(left-right)>1)
            isBalanced=false;
        return Math.max(left,right)+1;

    }
    public boolean isBalanced(TreeNode root) {
        dfs(root);
        return isBalanced;
    }
}