Explain how to delete a node from a doubly linked list given a pointer to that node.

• A. Relink neighbors: if node.prev != NULL, node.prev.next = node.next; if node.next != NULL, node.next.prev = node.prev; free node.
• B. Simply free the node without relinking neighbors.
• C. Set node.prev.next = NULL; node.next.prev = NULL; free node.
• D. Update only node.next and leave node.prev intact.

Answer: (A)

In a doubly linked list, deleting a node must remove it from the chain without leaving the rest of the list broken. You do this by relinking its neighbors: connect the previous node’s next pointer to the target’s next node, and connect the next node’s prev pointer to the target’s previous node. If either neighbor is absent (the node is at the head or tail), you skip that side safely. After the links are updated, free the target node. This preserves the continuity of the list and prevents any dangling references.

The method that relinks both sides before freeing is the right one because simply freeing the node leaves the neighboring nodes still pointing to a memory region that’s no longer part of the list, which can crash or corrupt the list. Setting one side to NULL or attempting to NULL out pointers without connecting the two neighbors would either truncate the list or leave broken links. Freeing without relinking, or truncating only part of the links, breaks the integrity of the structure.