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

class SingleLL :
def __init__(self) :

#  Add new node at the end of linked list
def addNode(self, value) :
# Create  node
if (self.head == None) :
else :
#  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) :
while (temp != None) :
#  Display node value
print("  ", temp.data, end = "")
#  Visit to next node
temp = temp.next

else :

def main() :
sll = SingleLL()
#  1 → 2 → 3 → 4 → 5 → 6 → 7 → 8 → NULL
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
def __init__(self, data) :
self.data = data
self.next = None

class SingleLL :
def __init__(self) :
self.tail = None

#  Add new node at the end of linked list
def addNode(self, value) :
#  Create a new node
if (self.head == None) :
else :
self.tail.next = node

self.tail = node

#  Display linked list element
def display(self) :
if (self.head == None) :
return

#  iterating linked list elements
while (temp != None) :
print("", temp.data ,"→", end = "")
#  Visit to next node
temp = temp.next

print(" NULL")

def main() :
sll = SingleLL()
#  10 → 20 → 30 → 40 → 50 → NULL
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).

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