637. Average of Levels in Binary Tree

1.问题描述

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.Example 1:Input:3/ \9  20/  \15   7Output: [3, 14.5, 11]Explanation:The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].Note:1. The range of node's value is in the range of 32-bit signed integer.来自 <https://leetcode.com/problems/average-of-levels-in-binary-tree/description/>

2.题目分析

求二叉树每一层的平均数，并存入到数组中，遍历方法使用层序遍历，关键是怎么判断一层结束了，区别于层序遍历的单队列，这里使用了两个队列，一个存储父节点层，另一个存储子节点层，再父节点层遍历完成后，计算均值并存储，然后交换这两个队列，使得当前子节点层成为下一次计算的父节点层。结束的条件是两个队列都空了。需要注意的是，当父结点层存在，子结点层不存在，即最后一层时，并没有计算最后一层均值，因此需要在循环结束时，计算最后一层均值并存储。

3.C++代码

   //我的代码：（beats 42%）vector<double> averageOfLevels(TreeNode* p){vector<double>r;queue<TreeNode*>q_root;queue<TreeNode*>q_child;q_root.push(p);int cnt = 0;double sum = 0;double aver = 0;while (!q_root.empty() || !q_child.empty()){if (!q_root.empty()){TreeNode *tmp = q_root.front();q_root.pop();if (tmp->left != NULL)q_child.push(tmp->left);if (tmp->right != NULL)q_child.push(tmp->right);cnt++;sum += tmp->val;}else{aver = sum / cnt;cnt = 0;sum = 0;q_root.swap(q_child);r.push_back(aver);}}aver = sum / cnt;r.push_back(aver);return r;}//改进版：（beats 74%）//只用一个队列，每次大循环处理一行，而不是只处理一个结点//利用每一层的结点个数就是队列的长度这一特点，减少一个队列vector<double> averageOfLevels2(TreeNode* p){vector<double>r;queue<TreeNode*>q;q.push(p);int n = 0;double sum = 0;double aver = 0;while (!q.empty()){n = q.size();sum = 0;for (int i = 0; i < n; i++){TreeNode*tmp = q.front();q.pop();sum += tmp->val;if (tmp->left)q.push(tmp->left);if (tmp->right)q.push(tmp->right);}aver = sum / n;r.push_back(aver);}return r;}

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637. Average of Levels in Binary Tree

1.问题描述

2.题目分析

3.C++代码

637. Average of Levels in Binary Tree相关推荐

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