Binary Tree Zigzag Level Order Traversal

description

Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Example 2:

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

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

submission

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn zigzag_level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        // alternating "queues"
        let (mut queue, mut swp) = (vec![], vec![]);
        let mut ans = vec![];
        if let Some(node) = root {
            queue.push(node);
        }
        // outer loop
        while !queue.is_empty() {
            let mut level = vec![];
            // we need to reverse the vec since vec is filo
            queue.reverse();
            // inner loop
            while let Some(node) = queue.pop() {
                let node = node.borrow();
                level.push(node.val);
                if let Some(left) = node.left.clone() {
                    swp.push(left);
                }
                if let Some(right) = node.right.clone() {
                    swp.push(right);
                }
            }
            if ans.len() % 2 == 1 {
                level.reverse();
            }
            std::mem::swap(&mut queue, &mut swp);
            ans.push(level);
        }
        ans
    }
}