Average of Levels in Binary Tree
Given the <code>root</code> of a binary tree, return <em>the average value of the nodes on each level in the form of an array</em>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0637.Average%20of%20Levels%20in%20Binary%20Tree/images/avg1-tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> 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].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0637.Average%20of%20Levels%20in%20Binary%20Tree/images/avg2-tree.jpg" style="width: 292px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,15,7]
<strong>Output:</strong> [3.00000,14.50000,11.00000]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
/**
 * 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) {
        queue<TreeNode*> q{{root}};
        vector<double> ans;
        while (!q.empty()) {
            int n = q.size();
            long long s = 0;
            for (int i = 0; i < n; ++i) {
                root = q.front();
                q.pop();
                s += root->val;
                if (root->left) {
                    q.push(root->left);
                }
                if (root->right) {
                    q.push(root->right);
                }
            }
            ans.push_back(s * 1.0 / n);
        }
        return ans;
    }
};