Balanced Binary Tree

Problem:

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.

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

Return false.

12/29/2017 update
Solution 1:
Brute force,

 /**  
  * Definition for a binary tree node.  
  * public class TreeNode {  
  *   int val;  
  *   TreeNode left;  
  *   TreeNode right;  
  *   TreeNode(int x) { val = x; }  
  * }  
  */  
 class Solution {  
   public boolean isBalanced(TreeNode root) {  
     if (root == null) return true;  
     return Math.abs(depth(root.left) - depth(root.right)) <= 1 &&  isBalanced(root.left)    && isBalanced(root.right);   
   }  
   private int depth(TreeNode root) {  
     if (root == null) return 0;  
     return Math.max(depth(root.left), depth(root.right)) + 1;  
   }  
 }  

Solution 2:
Depth first search, use -1 to represent not balanced case.

 /**  
  * Definition for a binary tree node.  
  * public class TreeNode {  
  *   int val;  
  *   TreeNode left;  
  *   TreeNode right;  
  *   TreeNode(int x) { val = x; }  
  * }  
  */  
 class Solution {  
   public boolean isBalanced(TreeNode root) {  
     return dfs(root) != -1;  
   }  
   private int dfs(TreeNode root) {  
     if (root == null) return 0;  
     int L = dfs(root.left);  
     if (L == -1) return -1;  
     int R = dfs(root.right);  
     if (R == -1) return -1;  
     return Math.abs(L - R) <= 1 ? Math.max(R, L) + 1 : -1;  
   }  
 }  

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思路：
一开始题意理解错了，我以为只看root，结果是看所有的点。没必要用result type，不用result type也很简单。