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Free A-Level Computer Science guide
Build the complete Node class from the workshop, link two values, then trace a loop that visits each node exactly once.
Content owner: Michael Print · Written for A-Level learners · Checked against official specifications
A singly linked list is a chain of nodes. Each node stores a value and a reference to the next node. The program keeps a reference to the first node (the head), and the last node's next reference is None.
To traverse it, start at the head. Print the current value, replace the current reference with the next one, and stop when the current reference becomes None.
The arrow is a reference, not the number stored in the node. In the code below, node1.setNextNode(node2) creates the link.
This follows the workshop's Node, setNextNode, getValue and getNextNode example. The recording used currNode != None; the runnable version uses the usual Python spelling, currNode is not None.
class Node:
def __init__(self, value):
self.__value = value
self.__nextNode = None
def setNextNode(self, nextNode):
self.__nextNode = nextNode
def getValue(self):
return self.__value
def getNextNode(self):
return self.__nextNode
node1 = Node(5)
node2 = Node(6)
node1.setNextNode(node2)
currNode = node1
while currNode is not None:
print(currNode.getValue())
currNode = currNode.getNextNode()Output: 5 followed by 6, each on its own line. After visiting 6, getNextNode() returns None and the loop ends.
Worked example
currNode = node1: current value is 5.currNode = currNode.getNextNode(). The current reference now points to node2.node2's next reference. That value is None.None.This 59-second captioned excerpt shows the loop: start at node1, print the current value, follow getNextNode(), and stop at None.
The full code above also includes the class and method definitions that lead up to this point in the workshop.
Use a current-node reference to visit every node. While it is not None, print its value and replace it with the next node. The reference moves from the node holding 5 to the node holding 6, then becomes None so the loop finishes.
Suppose the chain is 5 → 6 → 7 → None. To insert a node, connect the new node to its successor, then redirect the preceding node to the new node. To remove the node holding 6 from this chain, change the node holding 5 so its next reference points directly to the node holding 7. A separate Python variable may still refer to the removed node; it is simply no longer reachable from the list's head.
OCR's data-structures clarification includes creating, traversing, adding and removing data. The two-node program above concentrates on construction and traversal so each reference change is visible before you attempt the other operations.
These are original practice questions, not past-paper questions. Trace each reference before opening the suggested answer.
Using the full example above, state the output in order and explain why the loop stops after the second value.
Indicative answer: The loop prints 5, then 6 (1 mark each). After the second print, currNode becomes None (1). The while condition then becomes false and the loop stops (1).
The list currently contains 5 → 6 → None. Write two Python statements using the existing class to append a node containing 7. State what the traversal loop will print.
Indicative answer: Create node3 = Node(7) (1); call node2.setNextNode(node3) (1); the chain from node1 is now 5 → 6 → 7 → None (1); the same traversal loop prints 5, 6, 7 (1).
The Node class packages the value and next reference with methods for reading or changing them. Read the object-oriented programming guide for attributes, methods, self and Python's double-underscore convention.
These are independently written explanations and practice questions. CompSciTutoring.co.uk is not affiliated with or endorsed by an examination board. The marking guidance is indicative; always check the syllabus for your examination year.
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Use the free examples first, then explore one-to-one tuition if guided practice would help.