Insert node at end of linked list in python

Write a program which is create and add linked list node at the end (tail, last) position in python.

Suppose we are inserted the following (10, 20, 30 ,40 ,50) node in a sequence.

#  Python 3 Program for
#  Insert linked list element at end position set A

#  Linked list node
class LinkNode :
    def __init__(self, data) :
        self.data = data
        self.next = None
    

class SingleLL :
    def __init__(self) :
        self.head = None
    
    #  Add new node at the end of linked list
    def addNode(self, value) :
        # Create  node
        node = LinkNode(value)
        if (self.head == None) :
            self.head = node
        else :
            temp = self.head
            #  Find the last node
            while (temp.next != None) :
                #  Visit to next node
                temp = temp.next
            
            #  Add node at last position
            temp.next = node
        
    
    # Display all Linked List elements
    def display(self) :
        if (self.head != None) :
            temp = self.head
            while (temp != None) :
                #  Display node value
                print("  ", temp.data, end = "")
                #  Visit to next node
                temp = temp.next
            
        else :
            print("Empty Linked list")
        
    

def main() :
    sll = SingleLL()
    #  Linked list
    #  1 → 2 → 3 → 4 → 5 → 6 → 7 → 8 → NULL
    sll.addNode(1)
    sll.addNode(2)
    sll.addNode(3)
    sll.addNode(4)
    sll.addNode(5)
    sll.addNode(6)
    sll.addNode(7)
    sll.addNode(8)
    print(" Linked List ")
    sll.display()

if __name__ == "__main__": main()

 Linked List
  1  2  3  4  5  6  7  8

Time complexity of above program is O(n). We can optimize above algorithm using one extra pointer. Which is hold the reference of last node. Below are implementation of this logic.

#  Python 3 Program for
#  Insert linked list element at end position set B

#  Linked list node
class LinkNode :
    def __init__(self, data) :
        self.data = data
        self.next = None
    

class SingleLL :
    def __init__(self) :
        self.head = None
        self.tail = None
    
    #  Add new node at the end of linked list
    def addNode(self, value) :
        #  Create a new node
        node = LinkNode(value)
        if (self.head == None) :
            self.head = node
        else :
            self.tail.next = node
        
        self.tail = node
    
    #  Display linked list element
    def display(self) :
        if (self.head == None) :
            return
        
        temp = self.head
        #  iterating linked list elements
        while (temp != None) :
            print("", temp.data ,"→", end = "")
            #  Visit to next node
            temp = temp.next
        
        print(" NULL")
    

def main() :
    sll = SingleLL()
    #  Linked list
    #  10 → 20 → 30 → 40 → 50 → NULL
    sll.addNode(10)
    sll.addNode(20)
    sll.addNode(30)
    sll.addNode(40)
    sll.addNode(50)
    print(" Linked List ")
    sll.display()

if __name__ == "__main__": main()

 Linked List
 10 → 20 → 30 → 40 → 50 → NULL

Time complexity of above program is O(1).