If you are given only a reference to a node (not the head) and you know it is not the tail, how can you delete that node?

• A. Delete by removing the current node from the list using a head pointer.
• B. Copy the data from the next node into the current node, set current.next = current.next.next; delete the next node; O(1)
• C. Replace the current node with a new node carrying the same data.
• D. Set current = current.next; deallocate the old node.

Answer: (B)

When you can only reference the node to delete and you know it isn’t the tail, you don’t have access to the previous node to relink the list. The trick is to make the current node take on the identity of its next node and then remove that next node. Specifically, you copy the data from the next node into the current node, then bypass the next node by linking current.next to current.next.next, and finally delete the next node. This achieves deletion in constant time without requiring a head pointer, because you’re effectively removing the next node and leaving the current node with the next node’s value.

This approach works only if there is a next node to copy from; if the node were the tail, there is no next node to copy, and you would need access to the previous node or the head to perform deletion.