Python Program to Display all the Nodes in a Linked List using Recursion
Description:
To create a linked list and display the elements in a linked list. First, we need to create a Node class to store the data and address of the next Node. After that create a LinkedList class with a head instance. In LinkedList class, we have to define two methods one for placing the data in the linked list(create()) and another one to display the data of the link list (display()). In the first method(create()) we have to create a new Node and put the data into that node after that link to the previous node to the new node. And in display() method we have to traverse linked list recursively until 'node != None'. To traverse in linked list recursively we need an extra method to initialize recursion. Take look on the program for better understanding.
Python3 Program:
class Node:
def __init__(self,data):
self.data = data
self.next = None
class LinkList:
def __init__(self):
self.head = None
def create(self):
'''
function for creating linked list
'''
##taking user input
element=list(map(int,input('Enter elements of LinkedList:').split()))
## put the first value in to the head
self.head = Node(element[0])
temp = self.head
## put the all value in linked list
for i in element[1:]:
temp.next = Node(i)
temp = temp.next
def init_display(self,node):
if node is None:
return None
##displaying elemment untill temp is nut None
print(node.data,end=" ")
return self.init_display(node.next)
def display(self):
node=self.head
print('Element of LinkList is:', end='')
self.init_display(node)
def main():
l = LinkList() ##object of LinkList class
l.create() ##create the link list and put the element in link list
l.display() ##for display linked list
if __name__ == '__main__':
main()
Output:
Enter elements of LinkedList:5 4 2 8 7
Element of LinkList is:5 4 2 8 7
#####################################
Time complexity of this program is O(n).
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