222. Count Complete Tree Nodes (

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

Note:

<u style="box-sizing: border-box;">Definition of a complete binary tree from Wikipedia:</u>
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 2^h nodes inclusive at the last level h.

Example:

我的解法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        return (root==nullptr) ? 0:recurr(root);
    }
    int recurr(TreeNode* root) {
        int left_num = (root->left==nullptr) ? 0:recurr(root->left);
        int right_num = (root->right==nullptr) ? 0:recurr(root->right);
        return 1+left_num+right_num;
    }
};

Runtime: 40 ms, faster than 77.31% of C++ online submissions for Count Complete Tree Nodes.
Memory Usage: 31.4 MB, less than 84.20% of C++ online submissions for Count Complete Tree Nodes.

第一反应就是递归，但是发现速度并不是特别特别快，

递归的写法，看评论区：

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

但其实也是40ms， 不过用一个static的count variable还挺新奇的。

还有一行流

class Solution {
public:
    int countNodes(TreeNode* root) {
        return !root ? 0 : ((root->left ? countNodes(root->left) : 0) + (root->right ? countNodes(root->right) : 0) + 1);
    }
};

确实没想到调用自己！
但号称100%结果runtime 44ms