[LeetCode 解题报告]222. Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

Note:

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.

Example:

Input: 
    1
   / \
  2   3
 / \  /
4  5 6

Output: 6

Method 1. 

/**
 * 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 == NULL)
            return 0;
        
        int depthLeft = 0, depthRight = 0;
        TreeNode* curLeft = root, *curRight = root;
        while (curLeft) {
            depthLeft ++;
            curLeft = curLeft->left;
        }
        while (curRight) {
            depthRight ++;
            curRight = curRight->right;
        }
        
        if (depthLeft == depthRight)
            return pow(2, depthLeft) - 1;
        
        return countNodes(root->left) + countNodes(root->right) + 1;
    }
};

Method 2. 

/**
 * 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 == NULL)
            return 0;
        
        int res = 0;
        queue<TreeNode*> que({root});
        
        while (!que.empty()) {
            TreeNode* cur = que.front();
            res ++;
            que.pop();
            
            if (cur->left)
                que.push(cur->left);
            if (cur->right)
                que.push(cur->right);
        }
        return res;
    }
};