How do you delete a node from a doubly linked list given only a pointer to that node?

• A. If node.prev != null, node.prev.next = node.next; if node.next != null, node.next.prev = node.prev; free node; O(1)
• B. You must traverse from the head to find its predecessor
• C. Adjust only the previous node's next pointer
• D. Adjust only the next node's prev pointer

Answer: (A)

Deleting a node in a doubly linked list when you have a pointer to that node is done by splicing it out in constant time. Since every node has pointers to both its previous and next nodes, you can bypass the node by rewiring its neighbors: if there is a previous node, set that node’s next to the current node’s next; if there is a next node, set that node’s prev to the current node’s prev. Then free the current node. This is O(1) because you perform only a fixed number of pointer updates, with no traversal.

Edge cases are handled naturally: if the node is at the head, you skip updating a non-existent previous pointer and just update the next node’s prev to null; if it’s at the tail, you skip updating a non-existent next pointer and just update the previous node’s next to null. If you rely on an external head pointer, you may need to update that pointer externally when removing the head, but the internal splicing remains O(1).