LeetCode-Balanced Binary Tree

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.

关于平衡二叉树，一下是百度百科给出的说明：

平衡二叉树（Balanced Binary Tree）又被称为AVL树（有别于AVL算法），且具有以下性质：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。平衡二叉树的常用实现方法有红黑树、AVL、替罪羊树、Treap、伸展树等。 最小二叉平衡树的节点的公式如下 F(n)=F(n-1)+F(n-2)+1 这个类似于一个递归的数列，可以参考Fibonacci(斐波那契)数列，1是根节点，F(n-1)是左子树的节点数量，F(n-2)是右子树的节点数量。

伪代码：

IsHeightBalanced(tree)    return (tree is empty) or            (IsHeightBalanced(tree.left) and            IsHeightBalanced(tree.right) and            abs(Height(tree.left) - Height(tree.right)) <= 1)

代码如下：

/** * 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) {           return maxDepth(root) !=-1;    }        public int maxDepth(TreeNode root){        if(root == null)        return 0;     //树为空树，返回，此时为平衡二叉树        int left = maxDepth(root.left);        int right = maxDepth(root.right);        if(left == -1 || right == -1 ||Math.abs(left-right) > 1)        return -1;        return Math.max(left,right) + 1;    }}