# 637. Average of Levels in Binary Tree

#### Easy

***

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:**

![](https://assets.leetcode.com/uploads/2021/03/09/avg1-tree.jpg)

<pre><code>Input: root = [3,9,20,null,null,15,7]
<strong>Output:
</strong> [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].
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/03/09/avg2-tree.jpg)

<pre><code>Input: root = [3,9,20,15,7]
<strong>Output:
</strong> [3.00000,14.50000,11.00000]
</code></pre>

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[1, 104]`.
* `-231 <= Node.val <= 231 - 1`

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        if not root:
            return []
        END = "$"
        q = [root, END]
        result = []
        prev = None
        s = 0
        count = 0
        while q:
            node = q.pop(0)
            if node == END and prev == END:
                continue
            if node == END:
                result.append(s/count)
                count = 0
                s = 0
                q.append(END)
                prev = node
                continue
            count += 1
            s += node.val
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
            prev = node
        return result
        
```