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 7
Output: [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:
题目很好理解，就是求二叉树每一层的平均值。
我们可以创建一个用来存储每层节点值和的数组，和一个存储每层节点个数的数组，通过递归的方式来求二叉树的层平均值，注意在求当前层时要判断层数是否超过了数组的大小，一旦超过，就要扩大数组，一遍存储当前层的信息。
/** * 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: vector<double> averageOfLevels(TreeNode* root) { getAverage(root, res, 0); for(int i = 0; i < res.size(); ++i){ res[i] /= nums[i]; } return res; } void getAverage(TreeNode* root, vector<double> &res, int level){ if(root == nullptr) return; if(level >= res.size()){ res.push_back(0); nums.push_back(0); } res[level] += root->val; nums[level]++; getAverage(root->left, res, level+1); getAverage(root->right, res, level+1); } private: vector<double> res; vector<int> nums; };
LeetCode 637. Average of Levels in Binary Tree二叉树的层平均值 (C++)
原文：https://www.cnblogs.com/silentteller/p/10743503.html