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.

## Solution

### BFS Approach

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public IList<double> AverageOfLevels(TreeNode root) {
var list = new List<double>();
Queue<TreeNode> q = new Queue<TreeNode>();
q.Enqueue(root);
while(q.Count > 0)
{
int n = q.Count;
double sum = 0.0;
for(int i=0;i<n;i++)
{
TreeNode node = q.Dequeue();
sum += node.val;
if(node.left != null) q.Enqueue(node.left);
if(node.right != null) q.Enqueue(node.right);
}