題目連結

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

題解

思路和層次遍歷(點選檢視)一樣，沒什麼區別。

// 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) {
        // 空樹，返回空陣列
        if (root==nullptr)
            return this->result;
        // 父層節點，即當前正在遍歷的節點
        queue<TreeNode*> parentNodes;
        parentNodes.push(root);
        // 遍歷父層節點的同時獲取下一層（子層）節點
        while (!parentNodes.empty()){
            // 子層結點，即下一層節點
            queue<TreeNode*> childNodes;
            // 當前層的節點的值
            vector<double> valOfLevel;
            // 遍歷當前層
            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);
            }
            // 計算並存儲當前層節點值的平均值
            this->result.push_back(this->calcAverageOfLevel(valOfLevel));
            // 更新當前層
            parentNodes = childNodes;
        }

        return this->result;
    }
};
 

作者：@臭鹹魚

轉載請註明出處：https://www.cnblogs.com/chouxianyu/

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