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?

Analyse: root, root->left, root->right

1. Recursion

    Runtime: 0ms

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> result;
        preorder(root, result);
        return result;
    }
    void preorder(TreeNode* root, vector<int>& result){
        if(root){
            result.push_back(root->val);
            preorder(root->left, result);
            preorder(root->right, result);
        }
        return ;
    }
};

2. Iteration  

　 Runtime: 4ms

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> result;
        stack<TreeNode*> stk;
        if(root) stk.push(root);
        else return result;
        
        while(!stk.empty()){
            TreeNode* temp = stk.top();
            result.push_back(temp->val);
            stk.pop();
            
            if(temp->right) stk.push(temp->right);
            if(temp->left) stk.push(temp->left);
        }
        return result;
    }
};