[LC] 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,2,3]

Solution 1:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        stack, res = [], []
        if root is None:
            return res
        stack.append(root)
        while stack:
            cur = stack.pop()
            res.append(cur.val)
            if cur.right:
                stack.append(cur.right)
            if cur.left:
                stack.append(cur.left)
        return res

Solution 2: 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offerFirst(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.pollFirst();
            res.add(cur.val);
            if (cur.right != null) {
                queue.offerFirst(cur.right);
            }
            if (cur.left != null) {
                queue.offerFirst(cur.left);
            }
        }
        return res;
    }
}