LeetCode 94: Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization: The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> in = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        TreeNode node = root;
        
        while (!stack.isEmpty() || node != null)
        {
            if (node != null)
            {
                stack.push(node);
                
                // Left
                node = node.left;
            }
            else
            {
                node = stack.pop();
                
                // Center
                in.add(node.val);
                
                // Right
                node = node.right;
            }
        }
        
        return in;
    }
}