Leet Code OJ 110. Balanced Binary Tree [Difficulty: Easy]

题目：
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.

翻译：
给定一个二叉树，检测它是否是高度平衡的。
这个问题中，一个高度平衡的二叉树是定义在一个二叉树上，它的每个节点的两棵子树的高度相差都不超过1。

分析：
这其实就是判断是否是平衡二叉树。
平衡二叉树，又称AVL树。它或者是一棵空树，或者是具有下列性质的二叉树：它的左子树和右子树都是平衡二叉树，且左子树和右子树的高度之差之差的绝对值不超过1。
下面的做法，递归遍历每个节点，每次递归，都获取左右子树的高度进行判断，采用-1作为已经检测到不平衡的标志位，其余情况返回当前子树的高度，用作上一层递归判断。

代码：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public boolean isBalanced(TreeNode root) {int depth=getDepth(root,0);return depth!=-1;//-1代表子树出现不平衡}public int getDepth(TreeNode root,int length){if(root==null){return length;}int leftDepth=getDepth(root.left,length+1);int rightDepth=getDepth(root.right,length+1);if(leftDepth==-1||rightDepth==-1||leftDepth-rightDepth>1||rightDepth-leftDepth>1){return -1;}return leftDepth>rightDepth?leftDepth:rightDepth;}
}

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