Why can't you delete a node in a singly linked list if you are given only a pointer to that node, and not to its predecessor?

• A. Without the previous node you can't relink; a common trick is to copy data from the next node into the current and bypass the next, but you cannot delete the last node.
• B. You can delete it by updating its next pointer to NULL.
• C. You can delete by reassigning the node's memory.
• D. It's possible by deleting the current node and returning its value.

Answer: (A)

The key idea being tested is that removing a node from a singly linked list normally requires updating the previous node’s next pointer to skip over the deleted node. If you only have a pointer to the node itself and no access to its predecessor, you cannot relink the list to bypass that node.

A common technique to work around this limitation is to copy the data from the next node into the current node, then bypass the next node by setting current.next to current.next.next. This makes the list skip over the next node, and the current node ends up holding the next node’s data, so it looks like the current node was deleted. But this only works if there is a next node to copy from; if you’re dealing with the last node, there is no next node to copy, so you cannot perform the deletion with just a pointer to the current node.

That’s why the best answer points out that you can’t relink without the previous node, and notes the common data-copy trick while highlighting its limitation for the last node. The other options don’t fit because they either attempt to detach or delete the node without actually removing it from the list’s linkage, or rely on memory operations that don’t preserve the list structure.