Describe an in-place reversal of a doubly linked list and why it's simpler than reversing a singly linked list.

• A. Swap next and prev pointers for each node; do not update head and tail.
• B. Reverse by swapping data in each node, leaving the links unchanged.
• C. Swap next and prev pointers for each node, then swap the head and tail pointers.
• D. Rebuild a new list by detaching nodes and inserting them in reverse order.

Answer: (C)

The key idea is to use the two-way links that a doubly linked list already provides. Each node has both next and prev pointers, so you can reverse the list by swapping these two pointers for every node. After you’ve flipped the pointers on all nodes, the end that used to be the head becomes the tail and the end that used to be the tail becomes the head, so you also swap the list’s head and tail references. This is done in one pass and uses constant extra space because no new nodes are created or data is moved.

It’s simpler than reversing a singly linked list because, with a singly linked list, you only have a forward link and must carefully track the previous node to rewire the next pointers in reverse order, which requires extra bookkeeping. In the doubly linked case, the existing prev pointer lets you reverse direction directly, and you only need the final step of swapping head and tail to complete the reversal. Edge cases like an empty or single-element list are naturally handled by this approach. Rebuilding a new list or swapping only data would either use extra space or fail to truly reverse the node order.