Question
What will be the output of the following code when
printList is called with a linked list containing the values 1 -> 2 -> 3? class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def append(self, data): new_node = Node(data) if not self.head: self.head = new_node return last_node = self.head while last_node.next: last_node = last_node.next last_node.next = new_node def printList(self): current = self.head while current: print(current.data, end=" -> ") current = current.next print("None") ll = LinkedList() ll.append(1) ll.append(2) ll.append(3) ll.printList()Solution
In this code, a linked list is implemented where each Node contains a data field and a reference to the next node. The LinkedList class provides methods to append nodes and print the list. The append method adds nodes to the end of the list, and printList iterates through the list, printing each node's data followed by " -> ". When nodes with values 1, 2, and 3 are appended, the list becomes 1 -> 2 -> 3, and printList outputs 1 -> 2 -> 3 -> None. Why Other Options Are Wrong: B) 1 -> None: This option is incorrect because it implies that only 1 is in the list, ignoring the 2 and 3 that were appended. C) 2 -> 3 -> None: This option is incorrect because it suggests that the list starts with 2, which is not true since 1 was added first. D) 3 -> 2 -> 1 -> None: This option is incorrect as it represents a reverse order of the list, which does not occur when appending in the given code. E) None: This option is incorrect as it implies there is no output or that the list is empty, which is not the case.
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