# Python Program For Flattening A Linked List

Last Updated : 13 Jun, 2022

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) â€“ 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!

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