Leetcode: 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
    \
     2
    /
   3

return [3,2,1].

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

Solution:

Recursive:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) 
    {
        ArrayList<Integer> solution = new ArrayList<Integer>();
        postorderTraversal(root, solution);
        return solution;
    }
    
    public void postorderTraversal(TreeNode root, ArrayList<Integer> array)
    {
        if (root == null)
            return;
        postorderTraversal(root.left, array);
        postorderTraversal(root.right, array);
        array.add(root.val);
    }
}

Iterative:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) 
    {
        HashSet<TreeNode> used = new HashSet<TreeNode>();
        ArrayDeque<TreeNode> stack = new ArrayDeque<TreeNode>();
        ArrayList<Integer> solution = new ArrayList<Integer>();
        if (root!=null)   
        {
            stack.addFirst(root);
        }
        while(!stack.isEmpty())
        {
            TreeNode temp = stack.removeFirst();
            if(used.contains(temp))
                solution.add(temp.val);
            else
            {
                used.add(temp);
                stack.addFirst(temp);
                if(temp.right != null)
                    stack.addFirst(temp.right);
                if(temp.left != null)
                    stack.addFirst(temp.left);
            }
        }
        return solution;
    }
}