题目：给定一颗二叉树，用非递归方法中序遍历这颗二叉树

算法：

从某个节点node出发

1. 向左遍历，边遍历边将节点压入栈

2. 到达左子树的尽头，从栈中取出节点node并访问值

3. 转向node节点的右子树

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
       /**
		 * Algorithm:
		 * 
		 * 从某个节点node出发
		 *    1. 向左遍历，边遍历边将节点压入栈
		 *    2. 到达左子树的尽头，从栈中取出节点node并访问值
		 *    3. 转向node节点的右子树
		 */
	    public List<Integer> inorderTraversal(TreeNode root) {
	    	Stack<TreeNode> stack = new Stack<TreeNode>();
	    	List<Integer> inorderList = new ArrayList<Integer>();
	    
	    	TreeNode node = root;
	    	while (null!=node || !stack.empty()) {
	    		while (null != node) {
	    			stack.add(node);
	    			node = node.left;
	    		}
	    		if (!stack.empty()) {
	    			node = stack.pop();
	    			inorderList.add(node.val);
	    			node = node.right;
	    		}
	    	}
	    	return inorderList;
	    }
}