LeetCode 144:Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

 //递归方法：利用一个中间函数，将res作为输入参数
 class Solution {
 public:
	 vector<int> preorderTraversal(TreeNode* root) {
		 vector<int> res;
		 return tmpFunction(root, res);
	 }

	 vector<int> tmpFunction(TreeNode* root, vector<int> &res){
		 if (root == NULL) return res;
		 res.push_back(root->val);
		 tmpFunction(root->left,  res);
		 tmpFunction(root->right, res);
		 return res;
	 }
 };

 //非递归方法：
 //利用一个栈结构,现将头结点root压入栈中，然后从stack中弹出栈顶节点，记为tmp，然后将tmp的值压入vector<int> res中，
 //再将节点tmp的右孩子（不为空的话）先压入stack中，最后将tmp的左孩子（不为空的话）压入stack中，
 //不断重复以上操作，直到stack为空
 class Solution {
 public:
	 vector<int> preorderTraversal(TreeNode* root) {
		 vector<int> res;
		 stack<TreeNode*> s;
		 TreeNode* tmp;
		 if (root == NULL) return res;
		 else
		 {
			 s.push(root);
			 while (!s.empty())
			 {
				 tmp = s.top();
				 s.pop();
				 res.push_back(tmp->val);

				 if (tmp->right != NULL)  s.push(tmp->right);
				 if (tmp->left != NULL)    s.push(tmp->left);
			 }
		 }
		 return res;
	 }
 };