标签：exp cti array level xpl opened gif nat inpu
题目描述：
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:
解题思路：
使用queue，遍历树的每一层，确保每次queue中只保存了一层的节点。
代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<double> averageOfLevels(TreeNode* root) { 13 vector<double> ret; 14 queue<TreeNode*> nodes; 15 nodes.push(root); 16 while (!nodes.empty()) { 17 double sum = 0; 18 int size = nodes.size(); 19 for (int i = 0; i < size; ++i) { 20 TreeNode* tmp = nodes.front(); 21 nodes.pop(); 22 sum += tmp->val; 23 if (tmp->left) 24 nodes.push(tmp->left); 25 if (tmp->right) 26 nodes.push(tmp->right); 27 } 28 ret.push_back(sum / size); 29 } 30 return ret; 31 } 32 };
637. Average of Levels in Binary Tree
标签：exp cti array level xpl opened gif nat inpu
原文地址：https://www.cnblogs.com/gsz-/p/9520345.html
