## 637. Average of Levels in Binary Tree

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.

## Rust Solution

``````struct Solution;
use rustgym_util::*;

trait Preorder {
fn preorder(&self, level: usize, levels: &mut Vec<Vec<i32>>);
}

fn preorder(&self, level: usize, levels: &mut Vec<Vec<i32>>) {
if let Some(node) = self {
let node = node.borrow();
let left = &node.left;
let right = &node.right;
if levels.len() == level {
levels.push(vec![node.val]);
} else {
levels[level].push(node.val);
}
left.preorder(level + 1, levels);
right.preorder(level + 1, levels);
}
}
}

impl Solution {
fn average_of_levels(root: TreeLink) -> Vec<f64> {
let mut levels: Vec<Vec<i32>> = vec![];
root.preorder(0, &mut levels);
levels
.iter()
.map(|v| v.iter().map(|&x| x as f64).sum::<f64>() as f64 / v.len() as f64)
.collect()
}
}

#[test]
fn test() {
let root = tree!(3, tree!(9), tree!(20, tree!(15), tree!(7)));
let res: Vec<f64> = vec![3f64, 14.5, 11f64];
assert_eq!(Solution::average_of_levels(root), res);
}
``````

Having problems with this solution? Click here to submit an issue on github.