# Search in a Binary Search Tree Problem & Solution

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.

## C++ Solution

#pragma GCC optimize("Ofast")
#pragma GCC optimization("max-inline-insns-recursive-auto")

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

/**
* 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 == nullptr) {
return nullptr;
}

if (root->val == val) {
return root;
}

if (root->val > val) {

// This is tail recursion, so as good as iterative approach.
return searchBST(root->left, val);
} else {
return searchBST(root->right, val);
}
}
};