110. Balanced Binary Tree

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.

/**
 * 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 isBalancedHelper(root) != -1;
    }

    private int isBalancedHelper(TreeNode root){
        if (root == null) return 0;
        int left = isBalancedHelper(root.left);
        int right = isBalancedHelper(root.right);
        if (left == -1 || right== -1|| Math.abs(left - right) > 1) return -1;

        return Math.max(left, right) + 1; 
    }
}

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
# class Solution {
#     public boolean isBalanced(TreeNode root) {
#         return isBalancedHelper(root) != -1;
#     }

#     private int isBalancedHelper(TreeNode root){
#         if (root == null) return 0;
#         int left = isBalancedHelper(root.left);
#         int right = isBalancedHelper(root.right);
#         if (left == -1 || right== -1|| Math.abs(left - right) > 1) return -1;

#         return Math.max(left, right) + 1; 
#     }
# }
class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def helper(root):
            if not root: return 0

            left, right = helper(root.left), helper(root.right)
            if left == -1 or right == -1 or abs(left - right) > 1: return -1

            return max(left, right) + 1

        return helper(root) != -1