222. Count Complete Tree Nodes

Problem:

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.

Analysis:

This problem has nothing to do with binary search. The key is to compare the height of root's left and right tree. If left == right, it's simple, number of nodes is pow(2, h) - 1. Otherwise, countNodes(root.left) + countNodes(root.right) + 1

Solution:

Calculating height from root to left and only root.left are both ok. The latter one requires null check as follows.

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