Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

Return 0 / 1 ( 0 for false, 1 for true ) for this problem

Input : 
          1
         / \
        2   3

Return : True or 1 

Input 2 : 
         3
        /
       2
      /
     1

Return : False or 0 
         Because for the root node, left subtree has depth 2 and right subtree has depth 0. 
         Difference = 2 > 1. 

Method:

DFS

Solution:

Time: O(n)
Space: O(n)

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