Reverse LL (Recursive)

Given a linked list, reverse it using recursion.

You don't need to print the elements, just reverse the LL duplicates and return the head of updated LL.

Input format : Linked list elements (separated by space and terminated by -1)`

Sample Input 1 :

1 2 3 4 5 -1

Sample Output 1 :

5 4 3 2 1

	class Node:
	def __init__(self,data):
	self.data=data
	self.next=None
	def LinkedList(arr):
	head=None
	tail=None
	if len(arr)<1:
	return None
	else:
	for i in arr:
	if i==-1:
	break
	else:
	NewNode=Node(i)
	if head is None:
	head=NewNode
	tail=NewNode
	else:
	tail.next=NewNode
	tail=NewNode
	return head
	def printLL(head):
	while head is not None:
	print(head.data,end=" ")
	head=head.next
	print()
	def reverseLL(head):
	if head is None or head.next is None:
	return head,head
	smallHead,smallTail=reverseLL(head.next)
	smallTail.next=head
	head.next=None
	return smallHead,head
	# Main
	from sys import setrecursionlimit
	setrecursionlimit(11000)
	arr=list(map(int,input().split()))
	head,tail=reverseLL(LinkedList(arr))
	printLL(head)

view raw Reverse LL (Recursive).py hosted with ❤ by GitHub

	#using LL
	class Node:
	def __init__(self,data):
	self.data=data
	self.next=None
	def LinkedList(arr):
	head=None
	if len(arr)<1:
	return
	for i in arr:
	if i==-1:
	break
	else:
	NewNode=Node(i)
	NewNode.next=head
	head=NewNode
	return head
	def printLL(head):
	while head is not None:
	print(head.data,end=" ")
	head=head.next
	t=int(input())
	for i in range(t):
	arr=list(map(int,input().split()))
	printLL(LinkedList(arr))

view raw Reverse LL.py hosted with ❤ by GitHub