# binary tree preorder travelsal

## binary tree preorder travelsal

### easy problem

Basic skills. Recursive way is easy. Iterative way is not hard too. DFS and push the value to the result when visiting a point at the first time.

Recursive way.

`````` 1 /**
2  * Definition for binary tree
3  * struct TreeNode {
4  *     int val;
5  *     TreeNode *left;
6  *     TreeNode *right;
7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8  * };
9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode *root) {
13         vector < int > result;
14         result.clear();
15         preOrder(root,result);
16         return result;
17     }
18     void preOrder( TreeNode* root, vector < int >& result ) {
19         if ( !root ) {
20             return;
21         }
22         result.push_back(root->val);
23         preOrder(root->left,result);
24         preOrder(root->right,result);
25     }
26 };``````

Iterative way

`````` 1 /**
2  * Definition for binary tree
3  * struct TreeNode {
4  *     int val;
5  *     TreeNode *left;
6  *     TreeNode *right;
7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8  * };
9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode *root) {
13         vector < int > result;
14         result.clear();
15         if ( !root ) {
16             return result;
17         }
18         stack < TreeNode* > S;
19         while( !S.empty() ) {
20             S.pop();
21         }
22         S.push(root);
23         while( !S.empty() ) {
24             TreeNode* p = S.top();
25             S.pop();
26             result.push_back(p->val);
27             if ( p->right ) {
28                 S.push(p->right);
29             }
30             if ( p->left ) {
31                 S.push(p->left);
32             }
33         }
34         return result;
35     }
36 };``````