/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if(null == root) {
return true;
}
return Math.abs(height(root.left)  height(root.right)) < 2 && isBalanced(root.left) && isBalanced(root.right);
}
public int height(TreeNode root) {
if(null == root) {
return 0;
}
return 1 + Math.max(height(root.left), height(root.right));
}
}
Simple Java Recursive Solution O(nlogn)


I don't think time complexity is right.
This is my solution .
In first level, O(1) time to split T(n) to 2T(n/2)
second level, O(2) time to split 2T(n/2) to 4*(n/4)
Thus, time complexity is O(n).public class Solution { public boolean isBalanced(TreeNode root) { int res = helper(root); return res != 1; } private int helper(TreeNode root) { 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; } }

Complexity of Height() when called on a tree with 'n' nodes is O(n). (It visits every node once)
As we descend each level, Height() is called on subtrees with only n/2 nodes.Lets take a complete binary tree,
When isBalanced() calls Height() on the root node > n
When called from level 2 > n/2 + n/2
When called from level 3 > n/4 + n/4 + n/4 + n/4
......
When called from last level > again n (n/L x L times)So total complexity = n * no.of levels = O(nlogn)
PS: Just edit Height Function to remember when height difference of Child Subtrees > 1 for a O(n) solution.