144. Binary Tree Preorder Traversal

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?

Rust Solution

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);
}

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