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:

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].

```
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var averageOfLevels = function (root) {
if (root === null) {
return null;
}
let q = [root];
let averageArray = [];
while (q.length) {
let level = [];
let sum = 0;
let count = 0;
let loopTill = q.length;
// looping only till the length of q;
while (loopTill) {
// removing first element from queue
let node = q.shift();
// adding up node val to get sum & also incrementing count tokeep track of nodes at each level
sum = sum + node.val;
count = count + 1;
if (node.left !== null) {
level.push(node.left);
}
if (node.right !== null) {
level.push(node.right);
}
// reducing value as we have remove first element using shift() above
loopTill--;
}
//get average of this level
let averageOfLevel = sum / count;
averageArray.push(averageOfLevel);
q = level;
}
return averageArray;
};
```

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