110. Balanced Binary Tree 110.平衡二叉树

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 left and right subtrees of every node differ in height by no more than 1.

 

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

复习时还不会的地方：
为啥是max + 1呢，我觉得有时候是left + 1，有时候是right + 1
这个helper是每次+1的意思吗？是的吧，每次+1并且只+1，直到尽头。那肯定只有二者间较长的可以加了

一开始以为是最长路径和最短路径相减，不懂为啥直接是左右的最长路径相减了：
不是，只要任意一个|left - right| > 1就不算。所以随时准备返回false

然后还是用求二叉树最大长度的那个方法
有个奇葩的边界情况：左或者右节点为空时，长度返回-1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        //cc
        if (root == null) return true;
        return (helper(root) != -1);
    }
    
    public int helper(TreeNode root) {
        //cc
        if (root == null) return 0;
        
        int left = helper(root.left);
        int right = helper(root.right);
        
        if (left == -1 || right == -1 || Math.abs(left - right) > 1)
            return -1;
        
        return Math.max(left, right) + 1;
    }
}