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:
- The range of node's value is in the range of 32-bit signed integer.
以数组的形式返回每个层次上的节点的平均值
C++(19ms): bfs
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 if (root == NULL) 14 return vector<double>() ; 15 vector<double> res ; 16 queue<TreeNode*> que ; 17 que.push(root) ; 18 while(!que.empty()){ 19 long long sum = 0 ; 20 int len = que.size() ; 21 for(int i = 0 ; i < len ; i++){ 22 TreeNode* t = que.front() ; 23 que.pop() ; 24 sum += t->val ; 25 if (t->left != NULL) 26 que.push(t->left) ; 27 if (t->right != NULL) 28 que.push(t->right) ; 29 } 30 res.push_back(sum/(len*1.0)); 31 } 32 return res ; 33 } 34 };