Python dictionary with values as singly linked lists

Question

I have been trying to create a Python dictionary with values as singly linked lists. My code looks okay resulting correct keys. However each key yield the same value, do not know why. Please reply if you have any solution, code has been given below.

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 self.head is None:
            self.head = new_node
        else:
            tail = self.head
            while tail.next:
                tail = tail.next
            tail.next = new_node
    
    def create_two_ll(self, lists):
        ll_dict = {}
        for i in range(len(lists)):
            for data in lists[i]:
                self.append(data)
            ll_dict[f"ll{i}"] = ll
            print(ll_dict)
        print(ll_dict["ll0"].head.next.next.next.data)
        print(ll_dict["ll1"].head.next.next.next.data)
        return ll_dict

            
if __name__ == "__main__":
    ll = LinkedList()
    ll.create_two_ll([[1,3,5], [2,4,6]])

Output (wrong):

{'ll0': <__main__.LinkedList object at 0x7feead82c850>}
{'ll0': <__main__.LinkedList object at 0x7feead82c850>, 'll1': <__main__.LinkedList object at 0x7feead82c850>}
2
2

Answer 1

i solved it another way but im not sure its ok

i hope it'll trigger better way in your mind

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 self.head is None:
            self.head = new_node
        else:
            tail = self.head
            while tail.next:
                tail = tail.next
            tail.next = new_node
    
    def create_two_ll(self, lists):
        ll_dict = {}
        for i in range(len(lists)):
            for data in lists[i]:
                self.append(data)
            ll_dict[f"ll{i}"] = data
        print(ll_dict)
        return ll_dict

            
if __name__ == "__main__":
    ll = LinkedList()
    ll.create_two_ll([[1,3,5], [2,4,6],[7,9,11]])