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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        if (root==null){
            return new ArrayList<Integer>();
        }
        ArrayList<Integer> ret = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (!stack.empty()) {
            TreeNode p = null;
            //找到最左子树的叶子
            while (stack.peek() != null){
                p = stack.peek();
                stack.push(p.left);
                p = p.left;
            }
            //把最后叶子节点left的null弹出
            stack.pop();
            //如果是最右子树的叶子节点的时候栈就空了
            if (stack.empty()){
                return ret;
            }
            p = stack.pop();
            ret.add(p.val);
            //把右子树当做根节点，找最左
            stack.push(p.right);
        }
        return ret;
    }
}