222. 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.

直接遍历一遍会超时，代码如下：

 1 class Solution {
 2 public:
 3     int countNodes(TreeNode* root) {
 4         int count = 0;
 5         queue temp;
 6         if (root == NULL)
 7         {
 8             return 0;
 9         }
10         temp.push(root);
11         while (!temp.empty())
12         {
13             TreeNode * Node = temp.front();
14             count++;
15             temp.pop();
16             
17             if (Node->left != NULL)
18             {
19                 temp.push(Node->left);
20             }
21             if (Node->right != NULL)
22             {
23                 temp.push(Node->right);
24             }
25         }
26         return count;
27         }
28 };

 这道题求完全二叉树的结点的个数，所以对于一个结点，若是一直左子树遍历的个数和一直右子树遍历的个数相等，则说明是满二叉树，则计算公式为

2^h-1，若不相等则不是满二叉树，递归计算左子树和右子树再加上当前结点即可。

代码如下：

 1 class Solution {
 2 public:
 3     int countNodes(TreeNode* root) {
 4             int leftnum = 0;
 5             int rightnum = 0;
 6             TreeNode * ptr = root;
 7         
 8             if (root == NULL)
 9             {
10                 return 0;
11             }
12         
13             while (1)
14             {
15                 ptr = ptr->left;
16                 if (ptr != NULL)
17                 {
18                     ++leftnum;
19                 }
20                 else
21                 {
22                     break;
23                 }
24             }
25             ptr = root;
26             while (1)
27             {
28                 ptr = ptr->right;
29                 if (ptr != NULL)
30                 {
31                     ++rightnum;
32                 }
33                 else
34                 {
35                     break;
36                 }
37             }
38             if (leftnum == 0 && rightnum == 0)
39             {
40                 return 1;
41             }
42             if (leftnum == rightnum)
43             {
44                 return pow(2, leftnum + 1) - 1;
45             }
46             return 1 + countNodes(root->left) + countNodes(root->right);
47         }
48 };