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.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;

impl Solution {
    pub fn average_of_levels(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<f64> {
        let mut ans = vec![];
        let mut q = VecDeque::new();
        if let Some(root_node) = root {
            q.push_back(root_node);
        }
        while !q.is_empty() {
            let n = q.len();
            let mut s: i64 = 0;
            for _ in 0..n {
                if let Some(node) = q.pop_front() {
                    let node_borrow = node.borrow();
                    s += node_borrow.val as i64;
                    if let Some(left) = node_borrow.left.clone() {
                        q.push_back(left);
                    }
                    if let Some(right) = node_borrow.right.clone() {
                        q.push_back(right);
                    }
                }
            }
            ans.push((s as f64) / (n as f64));
        }
        ans
    }
}
