LeetCode-222 Count Complete Tree Nodes

题目描述

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.

题目大意

求一棵完全二叉树的结点个数。

示例

E1

解题思路

Solution1：

用一个队列保存二叉树每层的结点数，每pop一个结点便使得结果自加一。

Solution2：

 利用递归的特性，每次将将当前结点的左右子结点递归给函数，在叶结点处返回0 。

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

代码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) {
        long long res = 0;
        if(root == NULL)
            return res;
        queue<TreeNode*> level;
        // 将根结点入队列
        level.push(root);
        
        // 若当前结点的左右子节点存在，将其入队列，将当前结点出队列，结果加一
        while(!level.empty()) {
            TreeNode* tmp = level.front();
            if(tmp->left)
                level.push(tmp->left);
            if(tmp->right)
                level.push(tmp->right);
            level.pop();
            ++res;
        }
        
        return res;
    }
};

代码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) {
        // 若当前结点为叶节点，则返回0
        if(!root) return 0;
        // 查找当前结点的子结点是否都是满二叉树，若是则不用继续递归，直接返回结果
        int hl=0, hr=0;
        TreeNode *l=root, *r=root;
        while(l) {hl++;l=l->left;}
        while(r) {hr++;r=r->right;}
        if(hl==hr) return pow(2,hl)-1;
        // 若不是，则需要进一步进行递归
        return 1+countNodes(root->left)+countNodes(root->right);
    }
};