<p>You are given the <code>root</code> of a binary search tree (BST) and an integer <code>val</code>.</p>

<p>Find the node in the BST that the node&#39;s value equals <code>val</code> and return the subtree rooted with that node. If such a node does not exist, return <code>null</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg" style="width: 422px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [4,2,7,1,3], val = 2
<strong>Output:</strong> [2,1,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg" style="width: 422px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [4,2,7,1,3], val = 5
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 5000]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>7</sup></code></li>
	<li><code>root</code> is a binary search tree.</li>
	<li><code>1 &lt;= val &lt;= 10<sup>7</sup></code></li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
          if(root == NULL)
          {
            return NULL;
          }
          if(root->val == val)
          {
              return root; 
          }
        if(root->val> val)
        {
            return  searchBST(root->left,val);
        }
       else
        {
            return searchBST(root->right,val);
        }
    }
};
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