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?

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        if (root == null) return new ArrayList();
        Deque<TreeNode> stack = new ArrayDeque();
        stack.push(root);
        List<Integer> res = new ArrayList();
        while (!stack.isEmpty()) {
            TreeNode curr = stack.poll();
            res.add(curr.val);
            if (curr.right != null) stack.push(curr.right);
            if (curr.left != null) stack.push(curr.left);
        }
        return res;
    }
}