Get Node data of the Nth Node from the end

Published: January 4, 2017

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"""
 Get Node data of the Nth Node from the end.
 head could be None as well for empty list
 Node is defined as
"""
class Node(object):

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

'''return back the node data of the linked list in the below method.'''


def GetNode(head, position):

    if head is None or position is None:
        return

    pointer = head
    arr = []
    while pointer is not None and pointer.data is not None:
        #print (pointer.data)
        arr.append(pointer.data)
        if pointer.next is not None:
            pointer = pointer.next
        else :
            pointer = None

    arr.append(None)

    arr.reverse()

    if (len (arr) > position):
        print (arr [position])


n5 = Node ()
n4 = Node (4, n5)
n3 = Node (3, n4)
n2 = Node (2, n3)
n1 = Node (1, n2)

GetNode(n1, 1)

# output 4