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Code Single linked list

Find second last element of linked list in python

Python program for Find second last element of linked list. Here problem description and explanation.

#  Python 3 program for
#  Find the second last node of a linked list

#  Node of Linked List
class LinkNode :
	def __init__(self, data) :
		#  Set node value
		self.data = data
		self.next = None
	

class SingleLL :
	def __init__(self) :
		self.head = None
		self.tail = None
	
	#  Add new node at the end of linked list
	def insert(self, value) :
		#  Create a new node
		node = LinkNode(value)
		if (self.head == None) :
			self.head = node
		else :
			self.tail.next = node
		
		self.tail = node
	
	#  Display linked list element
	def display(self) :
		if (self.head == None) :
			return
		
		temp = self.head
		#  iterating linked list elements
		while (temp != None) :
			print(temp.data ," → ", end = "")
			#  Visit to next node
			temp = temp.next
		
		print("null")
	
	# Find the second last node of a linked list
	def secondLast(self) :
		node = self.head
		if (node == None) :
			print("Empty linked list")
		elif (node.next == None) :
			print("Only one node in this linked list")
		else :
			#  Find second last node
			while (node.next != None and node.next.next != None) :
				#  Visit to second next node
				node = node.next.next
			
			print("Second last element is : ", node.data)
		
	

def main() :
	sll = SingleLL()
	#  Add linked list node
	sll.insert(6)
	sll.insert(3)
	sll.insert(2)
	sll.insert(7)
	sll.insert(1)
	sll.insert(9)
	print("Linked List")
	#  6 → 3 → 2 → 7 → 1 → 9 → null
	sll.display()
	sll.secondLast()

if __name__ == "__main__": main()

Output

Linked List
6  → 3  → 2  → 7  → 1  → 9  → null
Second last element is :  1

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