Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},
1. 1
2. \
3. 2
4. /
5. 3

return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
Solution:
1. /*
2. * Definition for a binary tree node.
3. * public class TreeNode {
4. * int val;
5. * TreeNode left;
6. * TreeNode right;
7. * TreeNode(int x) { val = x; }
8. * }
9. /
10. public class Solution {
11. public List postorderTraversal(TreeNode root) {
12. List res = new ArrayList();
13. 14. if (root == null)
15. return res;
16. 17. Stack s1 = new Stack();
18. Stack s2 = new Stack();
19. 20. s1.push(root);
21. 22. while (!s1.isEmpty()) {
23. TreeNode node = s1.pop();
24. s2.push(node);
25. 26. if (node.left != null)
27. s1.push(node.left);
28. 29. if (node.right != null)
30. s1.push(node.right);
31. }
32. 33. while (!s2.isEmpty())
34. res.add(s2.pop().val);
35. 36. return res;
37. }
38. }