145. Binary Tree Postorder Traversal

Given the `root` of a binary tree, return the postorder traversal of its nodes' values.

Example 1:

```Input: root = [1,null,2,3]
Output: [3,2,1]
```

Example 2:

```Input: root = []
Output: []
```

Example 3:

```Input: root = [1]
Output: [1]
```

Example 4:

```Input: root = [1,2]
Output: [2,1]
```

Example 5:

```Input: root = [1,null,2]
Output: [2,1]
```

Constraints:

• The number of the nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`

Recursive solution is trivial, could you do it iteratively?

145. Binary Tree Postorder Traversal
``````struct Solution;
use rustgym_util::*;

trait Postorder {
fn postorder(&self, nodes: &mut Vec<i32>);
}

fn postorder(&self, nodes: &mut Vec<i32>) {
if let Some(node) = self {
let node = node.borrow();
node.left.postorder(nodes);
node.right.postorder(nodes);
nodes.push(node.val);
}
}
}

impl Solution {
fn postorder_traversal(root: TreeLink) -> Vec<i32> {
let mut res = vec![];
root.postorder(&mut res);
res
}
}

#[test]
fn test() {
let root = tree!(1, None, tree!(2, tree!(3), None));
let res = vec![3, 2, 1];
assert_eq!(Solution::postorder_traversal(root), res);
}
``````