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：

class Solution {
public:
    int countNodes(TreeNode* root) {
        int count = 0;
        queue<TreeNode *> temp;
        if (root == NULL)
        {
            return 0;
        }
        temp.push(root);
        while (!temp.empty())
        {
            TreeNode * Node = temp.front();
            count++;
            temp.pop();
            
            if (Node->left != NULL)
            {
                temp.push(Node->left);
            }
            if (Node->right != NULL)
            {
                temp.push(Node->right);
            }
        }
        return count;
        }
};

 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：

class Solution {
public:
    int countNodes(TreeNode* root) {
            int leftnum = 0;
            int rightnum = 0;
            TreeNode * ptr = root;
        
            if (root == NULL)
            {
                return 0;
            }
        
            while (1)
            {
                ptr = ptr->left;
                if (ptr != NULL)
                {
                    ++leftnum;
                }
                else
                {
                    break;
                }
            }
            ptr = root;
            while (1)
            {
                ptr = ptr->right;
                if (ptr != NULL)
                {
                    ++rightnum;
                }
                else
                {
                    break;
                }
            }
            if (leftnum == 0 && rightnum == 0)
            {
                return 1;
            }
            if (leftnum == rightnum)
            {
                return pow(2, leftnum + 1) - 1;
            }
            return 1 + countNodes(root->left) + countNodes(root->right);
        }
};