LeetCode OJ：Count Complete Tree Nodes（完全二叉树的节点数目）

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.
数二叉树的节点数首先肯定是可以用暴力揭发去解问题的，但是这样总是会timeout:

 1 class Solution {
 2 public:
 3     int countNodes(TreeNode* root) {
 4         if(!root) return 0;
 5         total++;
 6         countNodes(root->left);
 7         countNodes(root->right);
 8         return total;
 9     }
10 private:
11     int total;
12 };

其他的办法就是对完全二叉树而言，其一直向左走和一直向右边、走只可能是相同的或者相差1，如果相同那么起节点个数就是2^h - 1否曾再递归的加上左右节点的和就可以了。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int countNodes(TreeNode* root) {
13         if(!root) return 0;
14         TreeNode * leftNode = root;
15         TreeNode * rightNode = root;
16         int left = 0;
17         int right = 0;
18         while(leftNode->left){
19             left++;
20             leftNode=leftNode->left;
21         }
22         while(rightNode->right){
23             right++;
24             rightNode = rightNode->right;
25         }
26         if(left == right) return (1 << (left + 1)) - 1;
27         else return 1 + countNodes(root->left) + countNodes(root->right);
28     }
29 };

写的比较乱哈 ，凑合着看看把。