Merge two sorted linked lists in python

Python program for Merge two sorted linked lists. Here problem description and other solutions.

#  Python 3 program for
#  Merge two sorted lists

#  Linked list node
class LinkNode :
	def __init__(self, data) :
		self.data = data
		self.next = None
	

class SingleLL :
	def __init__(self) :
		self.head = None
		self.tail = None
	
	#  Add new node at end of linked list 
	def addNode(self, data) :
		node = LinkNode(data)
		if (self.head == None) :
			self.head = node
		else :
			#  Append the node at last position
			self.tail.next = node
		
		self.tail = node
	
	#  Display linked list element
	def display(self) :
		if (self.head == None) :
			print("\n Empty linked list")
			return
		
		temp = self.head
		# iterating linked list elements
		while (temp != None) :
			print("", temp.data , end = " →")
			#  Visit to next node
			temp = temp.next
		
		print(" NULL")
	
	#  Sorted Merge of two sorted list
	def mergeList(self, other) :
		if (other.head == None) :
			#  When have no element in second linked list
			return
		elif (self.head == None) :
			self.head = other.head
			self.tail = other.tail
		else :
			#  Define some useful resultant variable
			n1 = self.head
			n2 = other.head
			auxiliary = None
			#  Reset resultant list
			self.head = None
			self.tail = None
			#  iterate linked list element
			while (n1 != None or n2 != None) :
				if (n1 != None and n2 != None) :
					#  Compare front node element of both linked list
					if (n1.data <= n2.data) :
						#  Get front node of first linked list
						auxiliary = n1
						#  Visit to next node
						n1 = n1.next
					else :
						#  Get front node of second linked list
						auxiliary = n2
						#  Visit to next node
						n2 = n2.next
					
				elif (n1 != None) :
					auxiliary = n1
					#  Visit to next node
					n1 = n1.next
				else :
					auxiliary = n2
					#  Visit to next node
					n2 = n2.next
				
				if (self.head == None) :
					#  First node of resultant linked list
					self.head = auxiliary
					self.tail = auxiliary
				else :
					#  Add node at last of resultant list
					self.tail.next = auxiliary
					#  Set new last node
					self.tail = auxiliary
				
			
		
		other.head = None
		other.tail = None
	

def main() :
	#  Create linked list
	sll1 = SingleLL()
	sll2 = SingleLL()
	#  Sorted linked list 1
	#   1 → 7 → 8 → 15 → 19 → NULL
	sll1.addNode(1)
	sll1.addNode(7)
	sll1.addNode(8)
	sll1.addNode(15)
	sll1.addNode(19)
	#  Sorted linked list 2
	#   -2 → 5 → 6 → 12 → 16 → 18 → 19 → 31 → NULL
	sll2.addNode(-2)
	sll2.addNode(5)
	sll2.addNode(6)
	sll2.addNode(12)
	sll2.addNode(16)
	sll2.addNode(18)
	sll2.addNode(19)
	sll2.addNode(31)
	print("\n First Linked List")
	sll1.display()
	print(" Second Linked List")
	sll2.display()
	sll1.mergeList(sll2)
	#   -2 → 1 → 5 → 6 → 7 → 8 → 12 → 15 → 16 
	#   → 18 → 19 → 19 → 31 → NULL
	print(" After Merge")
	sll1.display()

if __name__ == "__main__": main()

Output

 First Linked List
 1 → 7 → 8 → 15 → 19 → NULL
 Second Linked List
 -2 → 5 → 6 → 12 → 16 → 18 → 19 → 31 → NULL
 After Merge
 -2 → 1 → 5 → 6 → 7 → 8 → 12 → 15 → 16 → 18 → 19 → 19 → 31 → NULL