Python Singlylinked list Element insertion Consider the foll

[Python] [Singly-linked list] [Element insertion] Consider the following singly linked list.

Provide the instructions to insert the new node immediately following the node containing 52. Do not use a loop or any additional external references.

Solution

# Node class
class Node:

# Function to initialise the node object
def __init__(self, data):
self.data = data # Assign data
self.next = None # Initialize next as null

class LinkedList:

# constructor to initialize head
def __init__(self):
self.head = None


# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node

def insert_After(self, prev_node, new_data):

# check if given prev_node exists
if prev_node is None:
print \"previous node not in the LinkedList.\"
return

# create new node and put the data
new_node = Node(new_data)

  
new_node.next = prev_node.next
prev_node.next = new_node

#function to print the linked list
def print_the_List(self):
temp = self.head
while (temp):
print temp.data,
temp = temp.next



# Code execution starts
if __name__==\'__main__\':

# Start with the empty list
list = LinkedList()



# Insert 36 at the beginning
list.push(36);
# Insert 18 at the beginning. So linked list becomes 18->36->None
list.push(18);
# Insert 52 at the beginning. So linked list becomes 52->18->36->None
list.push(52);
list.push(2);
list.push(73);
print \'The list is:\',
list.print_the_List()
# Insert 22, after 52. The linked list will becomes 73->2->22->52->18->36->None None
list.insert_After(list.head.next, 22)

print \'\ The list after inserting 22 is:\',
list.print_the_List()