# Average Of Levels In Binary Tree Problem

## Description

LeetCode Problem 637.

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

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Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000]
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].
``````

Example 2:

``````1
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Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]
``````

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• -2^31 <= Node.val <= 2^31 - 1

## Sample C++ Code

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
vector<double> res;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
long temp = 0;
int s = q.size();
for (int i = 0; i < s; i++) {
TreeNode* t = q.front();
q.pop();
if (t->left) q.push(t->left);
if (t->right) q.push(t->right);
temp += t->val;
}
res.push_back((double)temp/s);
}
return res;
}
};
``````