5.1.4 Binary Tree Inorder Traversal

Notes:
	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?
	Solution: 1. Iterative way (stack). Time: O(n), Space: O(n).
	2. Recursive solution. Time: O(n), Space: O(n).
	3. Threaded tree (Morris). Time: O(n), Space: O(1).
	*/
	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/

vector<int> inorderTraversal_1(TreeNode *root) {
        vector<int> res;
        stack<TreeNode *> stk;
        TreeNode *cur = root;
        while (cur || !stk.empty())
        {
            if (cur)
            {
                stk.push(cur);
                cur = cur->left;
            }
            else if (!stk.empty())
            {
                res.push_back(stk.top()->val);
                cur = stk.top()->right;
                stk.pop();
            }
        }
        return res;
    }