The property of BST (left < root < right) allows us to eliminate half of the tree in each recursive step.

1. Base case: If root is None or root.val == val, return root.
2. If val < root.val:
   • Search in the left subtree: return searchBST(root.left, val).
3. Else:
   • Search in the right subtree: return searchBST(root.right, val).

• Time Complexity: O(H) where H is tree height.
• Space Complexity: O(H).

def search_bst(root, val):
    if not root or root.val == val:
        return root
    
    if val < root.val:
        return search_bst(root.left, val)
    else:
        return search_bst(root.right, val)