LintCode Binary Tree Inorder Traversal 二叉树的中序遍历（非递归）

给出一棵二叉树,返回其中序遍历。
Given a binary tree, return the inorder traversal of its nodes’ values.

样例
给出二叉树 {1,#,2,3},
1
\
2
/
3
返回 [1,3,2].

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while(p != null || !stack.empty()) {
            while(p != null) {
                stack.push(p);
                p = p.left;
            }
            if(!stack.empty()) {
                p = stack.pop();
                list.add(p.val);
                p = p.right;
            }
        }
        return list;
    }
}