Node at a given index in linked list Python

PROGRAM TO GET Nth NODE IN A LINKED LIST

class Node:

    def __init__(self, data):

        self.data = data

        self.next = None

  

  

class LinkedList:

    def __init__(self):

        self.head = None

  

    ''' Given a reference (pointer to pointer) to the

        head of a list and an int, push a new node on

        the front of the list. '''

  

    def push(self, new_data):  # make new node and add

                              # into LinkedList

        new_node = Node(new_data)

        new_node.next = self.head

        self.head = new_node

  

    def getNth(self, llist, position):

  

        # call recursive method

        llist.getNthNode(self.head, position, llist)

  

    # recursive method to find Nth Node

    def getNthNode(self, head, position, llist):

        count = 0  # initialize count

        if(head):

            if count == position:  # if count is equal to position,

                                  # it means we have found the position

                print(head.data)

            else:

                llist.getNthNode(head.next, position - 1, llist)

        else:  # if head doesn't exist we have

              # traversed the LinkedList

            print('Index Doesn\'t exist')

  

  

# Driver Code

if __name__ == "__main__":

    llist = LinkedList()

    llist.push(1)

    llist.push(4)

    llist.push(1)

    llist.push(12)

    llist.push(1)

    # llist.getNth(llist,int(input()))

    # Enter the node position here

    # first argument is instance of LinkedList

  

    print("Element at Index 3 is", end=" ")

    llist.getNth(llist, 3)




OUTPUT

Element at index 3 is 4

CREDITS: https://www.geeksforgeeks.org/write-a-function-to-get-nth-node-in-a-linked-list/