题目链接

https://leetcode-cn.com/problems/average-of-levels-in-binary-tree/

题解

思路和层次遍历(点击查看)一样，没什么区别。

// Problem: LeetCode 637
// URL: https://leetcode-cn.com/problems/average-of-levels-in-binary-tree/
// Tags: Tree Queue
// Difficulty: Easy

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

struct TreeNode{
    int val;
    TreeNode* left;
    TreeNode* right;
};

class Solution{
private:
    vector<double> result;

    double calcAverageOfLevel(vector<double>& vec){
        double sum = 0;
        for (vector<double>::iterator it = vec.begin(); it!=vec.end(); ++it) sum += *it;
        return sum / vec.size();
    }

public:
    vector<double> averageOfLevels(TreeNode* root) {
        // 空树，返回空数组
        if (root==nullptr)
            return this->result;
        // 父层节点，即当前正在遍历的节点
        queue<TreeNode*> parentNodes;
        parentNodes.push(root);
        // 遍历父层节点的同时获取下一层（子层）节点
        while (!parentNodes.empty()){
            // 子层结点，即下一层节点
            queue<TreeNode*> childNodes;
            // 当前层的节点的值
            vector<double> valOfLevel;
            // 遍历当前层
            while (!parentNodes.empty()){
                root = parentNodes.front();
                parentNodes.pop();
                valOfLevel.push_back(root->val);
                if (root->left!=nullptr) childNodes.push(root->left);
                if (root->right!=nullptr) childNodes.push(root->right);
            }
            // 计算并存储当前层节点值的平均值
            this->result.push_back(this->calcAverageOfLevel(valOfLevel));
            // 更新当前层
            parentNodes = childNodes;
        }

        return this->result;
    }
};

作者：@臭咸鱼

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