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:

  1. The range of node's value is in the range of 32-bit signed integer.

代码:

 使用队列的先进先出的特点:

 1 class Solution {
 2 public:
 3     vector<double> averageOfLevels(TreeNode* root) {
 4         vector<double> ret;
 5         queue<TreeNode*> q;
 6         q.push(root);
 7         while (!q.empty()) {
 8             long long t = 0;
 9             int n = q.size();
10             for (int i = 0; i < n; i++) {
11                 auto node = q.front();
12                 q.pop();
13                 if (node->left) q.push(node->left);
14                 if (node->right) q.push(node->right);
15                 t += node->val;
16             }
17             ret.push_back(double(t)/double(n));
18         }
19         return ret;
20     }
21 };

 

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