Leetcode: Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Understand the problem:
First of all, what is a complete binary tree? 
A complete binary tree is a binary tree where each level except for the last level is full. 

A naive solution:
A naive solution is just to traverse the tree and count the number of nodes. The time complexity is O(n). It gets the Time Limit Exceeded. 

A Better Solution:
A better idea is to get the height of the left-most part, and height of the right-most part. If the left height and right height are the same, means the tree is full. Then the number of nodes is 2^h - 1. If not, we recursively count the number of nodes for the left sub-tree and right sub-tree. 

Code (Java):

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/**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution {     public int countNodes(TreeNode root) {         if (root == null) {             return 0;         }                   int leftHeight = findLeftHeight(root);         int rightHeight = findRightHeight(root);                   if (leftHeight == rightHeight) {             return (2 << (leftHeight - 1)) - 1;         }                   return countNodes(root.left) + countNodes(root.right) + 1;     }           private int findLeftHeight(TreeNode root) {         if (root == null) {             return 0;         }                   int height = 1;                   while (root.left != null) {             height++;             root = root.left;         }                   return height;     }           private int findRightHeight(TreeNode root) {         if (root == null) {             return 0;         }                   int height = 1;                   while (root.right != null) {             height++;             root = root.right;         }                   return height;     } }

The time complexity is O(h^2), because calculating the height of a binary tree takes O(h) time, and it recursively traverse the tree by O(h) time. 
In more detail, in worst case, we need to calculate the height of the tree in 1 + 2 + 3 + 4 + ... + h = O(h^2) time. It is actually 2 * O(h^2) because we need to calculate the left and right height.