Provide an outline for reversing a singly linked list iteratively.

• A. Initialize prev to head, curr to NULL, then traverse backward.
• B. Set curr to head; while curr is not NULL: next = curr.next; curr.next = head; head = curr; curr = next.
• C. Use a recursion that reverses sublists.
• D. Initialize prev to NULL, curr to head; while curr != NULL: next = curr.next; curr.next = prev; prev = curr; curr = next; head = prev.

Answer: (D)

Reversing a singly linked list iteratively hinges on flipping each node’s next pointer as you walk the list. Use two references: prev tracks the portion already reversed (initially NULL), and curr is the node you’re processing (initially the head). Before changing links, save the next node (next = curr.next). Then point the current node back to the previous one (curr.next = prev). Move forward by setting prev = curr and curr = next. Repeat until you’ve processed all nodes; at that moment, prev points to the new head, so head = prev. This works cleanly because you preserve access to the remainder of the list with next before re-linking, and you never lose nodes as you reverse the pointers. The other approaches either rely on a different setup (like starting with head as NULL) or attempt a recursive strategy, which isn’t the iterative method asked for, so they don’t fit the requirement.