### LeetCode Challenge #700. Search in a Binary Search Tree

You are given the `root` of a binary search tree (BST) and an integer `val`.

Find the node in the BST that the node’s value equals `val` and return the subtree rooted with that node. If such a node does not exist, return `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: []
```

Constraints:

• The number of nodes in the tree is in the range `[1, 5000]`.
• `1 <= Node.val <= 107`
• `root` is a binary search tree.
• `1 <= val <= 107`
##### Java Video Solution
###### C++ Solution
```				```
class Solution {
public:
TreeNode* searchBST(TreeNode* root, int val) {

if(!root)return NULL;

if(root->val==val)return root;
else if(root->val>val)return searchBST(root->left, val);
else return searchBST(root->right, val);
}
};
```
```
###### Java Solution
```				```
class Solution {
public TreeNode searchBST(TreeNode root, int val) {
TreeNode temp = root ;

while(temp!=null){
if(temp.val == val){
break;
}else if (temp.val > val){
temp = temp.left ;
}else{
temp = temp.right ;
}
}
return temp ;
}
}
```
```