Python Program For Flattening A Linked List
Last Updated :
23 Jul, 2025
Given a linked list where every node represents a linked list and contains two pointers of its type:
- Pointer to next node in the main list (we call it 'right' pointer in the code below).
- Pointer to a linked list where this node is headed (we call it the 'down' pointer in the code below).
All linked lists are sorted. See the following example
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
Write a function flatten() to flatten the lists into a single linked list. The flattened linked list should also be sorted. For example, for the above input list, output list should be 5->7->8->10->19->20->22->28->30->35->40->45->50.
The idea is to use the Merge() process of merge sort for linked lists. We use merge() to merge lists one by one. We recursively merge() the current list with the already flattened list.
The down pointer is used to link nodes of the flattened list.
Below is the implementation of the above approach:
Python3
# Python program for flattening
# a Linked List
class Node():
def __init__(self, data):
self.data = data
self.right = None
self.down = None
class LinkedList():
def __init__(self):
# Head of list
self.head = None
# Utility function to insert a
# node at beginning of the
# linked list
def push(self, head_ref, data):
# 1 & 2: Allocate the Node &
# Put in the data
new_node = Node(data)
# Make next of new Node as head
new_node.down = head_ref
# 4. Move the head to point to
# new Node
head_ref = new_node
# 5. Return to link it back
return head_ref
def printList(self):
temp = self.head
while(temp != None):
print(temp.data, end = " ")
temp = temp.down
print()
# An utility function to merge two
# sorted linked lists
def merge(self, a, b):
# If the first linked list is empty
# then second is the answer
if(a == None):
return b
# If second linked list is empty
# then first is the result
if(b == None):
return a
# Compare the data members of the
# two linked lists and put the
# larger one in the result
result = None
if (a.data < b.data):
result = a
result.down =
self.merge(a.down,b)
else:
result = b
result.down =
self.merge(a,b.down)
result.right = None
return result
def flatten(self, root):
# Base Case
if(root == None or
root.right == None):
return root
# Recur for list on right
root.right =
self.flatten(root.right)
# Now merge
root =
self.merge(root, root.right)
# Return the root
# It will be in turn merged with
# its left
return root
# Driver code
L = LinkedList()
'''
Let us create the following linked list
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
'''
L.head = L.push(L.head, 30);
L.head = L.push(L.head, 8);
L.head = L.push(L.head, 7);
L.head = L.push(L.head, 5);
L.head.right =
L.push(L.head.right, 20);
L.head.right =
L.push(L.head.right, 10);
L.head.right.right =
L.push(L.head.right.right, 50);
L.head.right.right =
L.push(L.head.right.right, 22);
L.head.right.right =
L.push(L.head.right.right, 19);
L.head.right.right.right =
L.push(L.head.right.right.right, 45);
L.head.right.right.right =
L.push(L.head.right.right.right, 40);
L.head.right.right.right =
L.push(L.head.right.right.right, 35);
L.head.right.right.right =
L.push(L.head.right.right.right, 20);
# Flatten the list
L.head = L.flatten(L.head);
L.printList()
# This code is contributed by maheshwaripiyush9
Output:
5 7 8 10 19 20 20 22 30 35 40 45 50
Time Complexity: O(N*N*M) – where N is the no of nodes in main linked list (reachable using right pointer) and M is the no of node in a single sub linked list (reachable using down pointer).
Space Complexity: O(N*M) as the recursive functions will use recursive stack of size equivalent to total number of elements in the lists.
Please refer complete article on Flattening a Linked List for more details!
