private int isBalancedUtil(Node root) {
        if (root == null) {
            return 0;
        }
        int left = isBalancedUtil(root.left);
        if (left < 0) {
            return -1;
        }
        int right = isBalancedUtil(root.right);
        if (right < 0) {
            return -1;
        }
        int diff = Math.abs(left - right);
        return diff <= 1 ? (Math.max(left, right) + 1) : -1;
    }
Date 10/06/2016
@author Tushar Roy

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

Time complexity O(logn)

Reference
https://leetcode.com/problems/balanced-binary-tree/
src/com/interview/tree/HeightBalanced.java