110. Balanced Binary Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/lixq05/article/details/79210353

本文介绍了两种判断二叉树是否高度平衡的方法：一种是递归检查左右子树的平衡状态及深度差；另一种是在发现不平衡时直接返回特定值。通过这两种方法可以有效地解决这一经典算法问题。

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.

解题思路（比较容易想到的）

左右子树都是平衡的以及左右子树的深度不超过1—得到根节点的树也是平衡的

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        //每个根节点的左右子树都是平衡树
        if(root==null)
            return true;
        //左子树和右子树都是平衡树 + 左右子树的深度不超过1
        if(isBalanced(root.left)&& isBalanced(root.right)){
            int l = getDepth(root.left);
            int r = getDepth(root.right);
            if(Math.abs(l-r)<=1)
                return true;
        }
        return false;
    }
    public int getDepth(TreeNode root){
        if(root==null)
            return 0;
        else{
            int left = getDepth(root.left);
            int right = getDepth(root.right);
            return Math.max(left,right)+1;
        }
    }
}

第二种：不平衡的子树就返回-1，否则返回其长度

class Solution {
    public boolean isBalanced(TreeNode root) {
      return maxDepth(root)!=-1;
    }
    public int maxDepth(TreeNode root){
        if(root==null)
            return 0;
        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        if(left==-1 || right==-1 || Math.abs(left-right)>1)
            return -1;
        else
            return Math.max(left,right)+1;

    }
}