700. Search DFS method in binary search tree

700. Search in binary search tree

Given a binary search tree （BST） The root node root And an integer value val.

You need to BST The node value found in is equal to val The node of . Return the subtree with this node as the root . If the node doesn't exist , Then return to null .

Example 1:

Input ：root = [4,2,7,1,3], val = 2
Output ：[2,1,3]

Example 2:

Input ：root = [4,2,7,1,3], val = 5
Output ：[]

The solution code is as follows ：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */
 struct TreeNode* re;

 void dfs(struct TreeNode* root,int val){
    
     if(root&&re==NULL){
    
         if(root->val==val){
    
             re=root;
         }
         dfs(root->left,val);
         dfs(root->right,val);
     }
 }


struct TreeNode* searchBST(struct TreeNode* root, int val){
    


    re=NULL;
    dfs(root,val);
    return re;


}