Binary Tree Preorder Traversal | Leetcode

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

---

Solution: Use a stack to hold the nodes. After popping each node, first add the value to the answer, then push the right node, then push the left node.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        if (root == null) return Collections.emptyList();

        List<Integer> answer = new ArrayList<>();
        LinkedList<TreeNode> stack = new LinkedList<>();
        stack.push(root);

        do {
            TreeNode node = stack.pop();
            answer.add(node.val);
            if (node.right != null) stack.push(node.right);
            if (node.left != null) stack.push(node.left);
        } while (!stack.isEmpty());

        return answer;
    }
}