Binary Tree Inorder Traversal - Leetcode

/**
 * 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) {
        Stack<TreeNode> s = new Stack<>();
		TreeNode pointer = root;
		ArrayList<Integer> ai = new ArrayList<>();
		while(pointer!=null || !s.empty()){
			if(pointer!=null){
				s.push(pointer);
				pointer = pointer.left;
			}else{
				ai.add(s.peek().val);
				pointer = s.pop().right;
			}
		}
		return ai;
    }
}

思路不难，了解就好：1）找到最左边节点，推入链表。栈来辅助中间走过的节点

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?