[FIXED] Status of ''self.head'' in a singly link list

Issue

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)

Solution

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