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public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;
        Stack<TreeNode> s = new Stack<TreeNode>();
        Stack<TreeNode> output = new Stack<TreeNode>();
        s.push(root);
        while(!s.isEmpty()){
            TreeNode curr = s.pop();
            output.push(curr);
            if(curr.left != null)
                s.push(curr.left);
            if(curr.right != null)
                s.push(curr.right);
        }
        while(!output.isEmpty()){
            res.add(output.pop().val);
        }
        return res;
    }
}

/**
 * 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?
 */
 
 /**
 * 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> res = new ArrayList<Integer>();
        if(root == null)
            return res;
        Stack<TreeNode> s = new Stack<TreeNode>();
        s.push(root);
        TreeNode prev = null;
        while(!s.isEmpty()){
            TreeNode curr = s.peek();
            if(prev == null || prev.left == curr || prev.right == curr){
                // traverse from top to bottom, and if curr has left child or right child, push into the stack; otherwise, pop out. 
                if(curr.left != null)
                    s.push(curr.left);
                else if(curr.right != null)
                    s.push(curr.right);
            }else if(curr.left == prev){
                if(curr.right != null)
                    s.push(curr.right);
            }else{
                res.add(curr.val);
                s.pop();
            }
            prev = curr;
        }
        return res;
    }
 }