Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [1,2,3]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Example 4:
Input: root = [1,2] Output: [1,2]
Example 5:
Input: root = [1,null,2] Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
struct Solution;
trait Preorder {
fn preorder(&self, all: &mut Vec<i32>);
}
impl Preorder for TreeLink {
fn preorder(&self, all: &mut Vec<i32>) {
if let Some(node) = self {
let node = node.borrow();
all.push(node.val);
node.left.preorder(all);
node.right.preorder(all);
}
}
}
impl Solution {
fn preorder_traversal(root: TreeLink) -> Vec<i32> {
let mut res = vec![];
root.preorder(&mut res);
res
}
}
#[test]
fn test() {
let root = tree!(1, None, tree!(2, tree!(3), None));
let res = vec![1, 2, 3];
assert_eq!(Solution::preorder_traversal(root), res);
}