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.

思路：判断是否是平衡二叉树，dfs即可，递归判断每个子树是否平衡二叉树，如果有子树不满足，则整体也不满足。

具体代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * 判断是否平衡二叉树
     * 看左右子树高度差是否超过1
     * 不超过1则分别判断左右子树是否平衡二叉树
     */ 
    public boolean isBalanced(TreeNode root) {
        if(root == null)
            return true;
        if(dep(0,root) > - 10)
        	return true;
        return false;
    }
    //计算树的高度
    private int dep(int dep,TreeNode root){
        if(root == null){
            return dep-1;
        }
        int dep1 = dep(dep+1,root.left);
        int dep2 = dep(dep+1,root.right);
        //如果高度差超过1,返回-10
        //则以后的dep返回值均为-10
        if(Math.abs(dep1 - dep2) > 1){
        	return -10;
        }
        return dep1 > dep2 ? dep1 : dep2;
    }
}