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 0;
        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 0;
        TreeNode * leftNode = root;
        TreeNode * rightNode = root;
        int left = 0;
        int right = 0;
        while(leftNode->left){
            left++;
            leftNode=leftNode->left;
        }
        while(rightNode->right){
            right++;
            rightNode = rightNode->right;
        }
        if(left == right) return (1 << (left + 1)) - 1;
        else return 1 + 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
    }
}