If you only have access to a head pointer, removing the last element requires traversing to which node?

• A. The second-to-last node
• B. The last node
• C. The first node
• D. The node after the head

Answer: (A)

In a singly linked list with only a head pointer, you don’t have a direct reference to the node before the last, so you must walk from the head to locate that predecessor. You keep moving until you reach the node whose next pointer refers to the last node (i.e., current.next.next is null). That node is the second-to-last. Once you have it, you unlink the last node by setting its next to null and freeing the last node. If the list has only one element, there’s no second-to-last to find, so you simply set the head to null. This is why removing the last element requires traversing to the second-to-last node.