leetCode 110.Balanced Binary Tree （平衡二叉树） 解题思路和方法

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;    }}