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 2hnodes inclusive at the last level h.

FIND A MORE EFFICIENT CODE!!!

Analyse: 

1. do level traversal, count and add up the node(s) at each level.

    NA:  TIME LIMIT EXCEEDED

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if(!root) return 0;
        queue<TreeNode* > qu;
        qu.push(root);
        int result = 0;
        while(!qu.empty()){
            int n = qu.size();
            result += n;
            while(n--){
                TreeNode* temp = qu.front();
                qu.pop();
                if(temp->left) qu.push(temp->left);
                if(temp->right) qu.push(temp->right);
            }
        }
        return result;
    }
};

2. Continuously judge whether the left level and right level are equal. If they are equal, return 2^level - 1; If not, compute 1 + nodes(left) + nodes(right).

    Runtime: 164ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if(!root) return 0;
        
        int leftLevel = 1, rightLevel = 1;
        TreeNode* temp = root->left;
        while(temp){
            leftLevel++;
            temp = temp->left;
        }
        temp = root->right;
        while(temp){
            rightLevel++;
            temp = temp->right;
        }
        
        if(leftLevel == rightLevel) 
            return (1 << leftLevel) - 1;
        
        return 1 + countNodes(root->left) + countNodes(root->right);
    }
};