ITPEC FE Subject B April 2024 Question 10

Inserting into a circular linked list by traversing to the tail node.

In a circular linked list, the last element's next points back to listHead. To insert after the last element, we must first find the tail — the node whose next equals listHead.

Finding the tail (blanks A and B):

The while loop must stop when we reach the node pointing back to the head:

• •A = `listHead`: the condition while (tmp.next is not listHead) traverses until we find the tail node
• •B = `tmp.next`: inside the loop, tmp ← tmp.next advances to the next node

After the loop, tmp is the last element. Then:
- tmp.next ← newNode — tail now points to the new node
- newNode.next ← listHead — new node points to head, closing the circle

Why not others:
- (a) A=newNode, B=listHead.next — the loop condition tmp.next is not newNode makes no sense since newNode was just created and isn't in the list yet; listHead.next doesn't traverse the list
- (c) A=tmp, B=listHead — tmp.next is not tmp is a meaningless self-comparison; tmp ← listHead resets instead of advancing
- (d) A=listHead, B=newNode.next — correct loop condition, but tmp ← newNode.next doesn't advance through the list since newNode isn't linked yet

Key rule: In a circular linked list, the tail node is identified by tail.next == head.

AI-generated — may contain errors