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解法去解问题的，但是这样总是会timeout:

 class Solution {
 public:
     int countNodes(TreeNode* root) {
         if(!root) return ;
         total++;
         countNodes(root->left);
         countNodes(root->right);
         return total;
     }
 private:
     int total;
 };

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。

 /**
  * 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 ;
         TreeNode * leftNode = root;
         TreeNode * rightNode = root;
         int left = ;
         int right = ;
         while(leftNode->left){
             left++;
             leftNode=leftNode->left;
         }
         while(rightNode->right){
             right++;
             rightNode = rightNode->right;
         }
         if(left == right) return ( << (left + )) - ;
         else return  + countNodes(root->left) + countNodes(root->right);
     }
 };

写的比较乱哈 ，凑合着看看把。

下面是java代码，首先肯定是递归形式的，不出所料，同样会TLE：

 /**
  * 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;
         else
             return 1 + countNodes(root.left) + countNodes(root.right);
     }
 }

考虑到完全二叉树的性质，下面这个写法就不会TLE了：

 public class Solution {
     public int countNodes(TreeNode root) {
         if(root == null)
             return 0;
         int right = 1, left = 1;
         TreeNode leftNode = root;
         TreeNode rightNode = root;
         while(leftNode.left != null){
             leftNode = leftNode.left;
             left++;
         }
         while(rightNode.right != null){
             rightNode = rightNode.right;
             right++;
         }
         if(left == right)
             return (1 << left) - 1;
         else
             return 1 + countNodes(root.left) + countNodes(root.right); //注意这里不要忘记计算本身的这个节点
     }
 }