How do you insert a node at the end of a circular singly linked list given a tail pointer?

• A. tail.next = new; new.next = head; tail = new; (If the list is empty, new.next = new; tail = new.)
• B. tail.next = head; head = new; tail = new;
• C. tail = new; new.next = head; head.next = new;
• D. head.next = new; new.next = head; tail = head;

Answer: (C)

In a circular singly linked list with a tail pointer, tail.next always points to the head of the list. To insert a node at the end, create the new node first. If the list is empty, make the new node point to itself and set the tail to the new node. If it’s not empty, link the new node after the current tail and then move the tail to the new node: set the new node’s next to tail.next (the head), set tail.next to the new node, and finally update tail to the new node. This keeps the circle intact and ensures tail remains the last node.

The reason this approach works is that it preserves the direct connection from tail to the head via tail.next, while inserting the new node right after tail and then updating tail to reflect the new end of the list. If you try to reassign tail before linking, or alter head’s next pointers in the wrong order, you risk breaking the circular link and losing part of the list.