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;
}
}