1. Each comparison moves one level down, so the cost equals the height of the tree.
2. A balanced BST has height about log n, but inserting keys in sorted order gives every node only a right child.
3. That degenerate chain has height n − 1, so the worst case is O(n).

Answer: O(n).

An unbalanced BST is just a linked list with extra pointers. Self-balancing variants such as AVL and red-black trees exist precisely to keep the height at O(log n).

Shortcut. Worst case means worst shape: keys inserted in sorted order make a chain of height n − 1, so a search can take O(n).

Common trap. "O(log n)" is the balanced or average case, guaranteed only by self-balancing trees such as AVL, not by a plain BST.