Python Program For Flattening A Linked List

Given a linked list where every node represents a linked list and contains two pointers of its type: 

1. Pointer to next node in the main list (we call it ‘right’ pointer in the code below).
2. 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!