Status of ''self.head'' in a singly link list

Question

I’ve been practicing link list but can’t understand what ”self.head” actually refers.
Is it the first value in a list at index 0? And how can I print data inside the head?

class Node:
    def __init__(self, data=None, next=None):
        self.data = data
        self.next = next


class Linkedlist:
    def __init__(self):
        self.head = None
    def print_var(self):
        itr = self.head
        print(itr.data)
def insert_at_begining(self, data):
    node = Node(data, self.head)
    self.head = node
if __name__ = '__main__':
ll = Linkedlsit()
ll.insert_at_begining(3)
ll.insert_at_begining(4)
ll.insert_at_begining(6)
ll.insert_at_begining(8)
ll.print()

If I’m to call print fuction function, it will through an error.
(Say, linklist is not empty)

Asked By: Aleef

||

Source

Answer 1

self.head refers to the first Node in the list. From there, you can "walk" the list with something like:

current_node = ll.head
while current_node:
    print(f"Current node data: {current_node.data}")
    current_node = current_node.next

print("Reached End of List")

Answered By: Cargo23

Answer 2

Check out my code and related comments

ll = LinkedList() # Create empty list
print(ll.head) # This prints None, because the list is empty, there are no nodes
ll.print_var() # This gives you an error. You're trying to print ll.head.data, but as shown before ll.head is None, not a Node
ll.head = Node('hello', ll.head) # The list now has one node
ll.print_var() # This is now working, and prints 'hello'

Answered By: Riccardo Bucco

Answer 3

After fixing your misspellings and calling the function by its proper name, print_var, it functions as it should. If you wanted to display your entire linked list you can do so with one of these methods:

    def __iter__(self):
        node = self.head
        while node is not None:
            yield node.data
            node = node.next
        yield node

    def show(self):
        node = self.head
        while node is not None:
            print(node.data, end=" -> ")
            node = node.next
        print(node)

__iter__ makes it an iterable you can loop through and show() will print the contents of the linked list.

As for self.head, it points to the newest node added to your linked list or None in an empty linked list.

Answered By: Sculby

Status of ''self.head'' in a singly link list

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