Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.
Example:
Input: [1,2,3,4,5] 1 / \ 2 3 / \ 4 5 Output: [[4,5,3],[2],[1]]
Explanation:
1. Removing the leaves [4,5,3] would result in this tree:
1
/
2
2. Now removing the leaf [2] would result in this tree:
1
3. Now removing the leaf [1] would result in the empty tree:
[][[3,5,4],[2],[1]], [[3,4,5],[2],[1]], etc, are also consider correct answers since per each level it doesn't matter the order on which elements are returned.
struct Solution;
use rustgym_util::*;
trait Postorder {
fn postorder(&self, leaves: &mut Vec<Vec<i32>>) -> usize;
}
impl Postorder for TreeLink {
fn postorder(&self, leaves: &mut Vec<Vec<i32>>) -> usize {
if let Some(node) = self {
let left = node.borrow().left.postorder(leaves);
let right = node.borrow().right.postorder(leaves);
let level = left.max(right) + 1;
if leaves.len() < level {
leaves.push(vec![node.borrow().val]);
} else {
leaves[level - 1].push(node.borrow().val);
}
level
} else {
0
}
}
}
impl Solution {
fn find_leaves(root: TreeLink) -> Vec<Vec<i32>> {
let mut res: Vec<Vec<i32>> = vec![];
root.postorder(&mut res);
res
}
}
#[test]
fn test() {
let root = tree!(1, tree!(2, tree!(4), tree!(5)), tree!(3));
let res = vec![vec![4, 5, 3], vec![2], vec![1]];
assert_eq!(Solution::find_leaves(root), res);
}