• Difficulty Level : Hard
• Last Updated : 17 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:

## C++

 `// C++ program for flattening a Linked List` `#include ` `using` `namespace` `std;`   `// Link list node ` `class` `Node` `{` `    ``public``:` `        ``int` `data;` `        ``Node *right, *down;` `};`   `Node* head = NULL;`   `// An utility function to merge two sorted` `// linked lists` `Node* merge(Node* a, Node* b)` `{` `    `  `    ``// If first linked list is empty then second` `    ``// is the answer` `    ``if` `(a == NULL)` `        ``return` `b;`   `    ``// If second linked list is empty then first` `    ``// is the result` `    ``if` `(b == NULL)` `        ``return` `a;`   `    ``// Compare the data members of the two linked ` `    ``// lists and put the larger one in the result` `    ``Node* result;`   `    ``if` `(a->data < b->data) ` `    ``{` `        ``result = a;` `        ``result->down = merge(a->down, b);` `    ``}`   `    ``else` `    ``{` `        ``result = b;` `        ``result->down = merge(a, b->down);` `    ``}` `    ``result->right = NULL;` `    ``return` `result;` `}`   `Node* flatten(Node* root)` `{` `    `  `    ``// Base Cases` `    ``if` `(root == NULL || root->right == NULL)` `        ``return` `root;`   `    ``// Recur for list on right` `    ``root->right = flatten(root->right);`   `    ``// Now merge` `    ``root = merge(root, root->right);`   `    ``// Return the root` `    ``// it will be in turn merged with its left` `    ``return` `root;` `}`   `// Utility function to insert a node at` `// beginning of the linked list` `Node* push(Node* head_ref, ``int` `data)` `{` `    `  `    ``// Allocate the Node & Put in the data` `    ``Node* new_node = ``new` `Node();`   `    ``new_node->data = data;` `    ``new_node->right = NULL;`   `    ``// Make next of new Node as head` `    ``new_node->down = head_ref;`   `    ``// Move the head to point to new Node` `    ``head_ref = new_node;`   `    ``return` `head_ref;` `}`   `void` `printList()` `{` `    ``Node* temp = head;` `    ``while` `(temp != NULL)` `    ``{` `        ``cout << temp->data << ``" "``;` `        ``temp = temp->down;` `    ``}` `    ``cout << endl;` `}`   `// Driver code` `int` `main()` `{` `    `  `    ``/* 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` `    ``*/` `    ``head = push(head, 30);` `    ``head = push(head, 8);` `    ``head = push(head, 7);` `    ``head = push(head, 5);`   `    ``head->right = push(head->right, 20);` `    ``head->right = push(head->right, 10);`   `    ``head->right->right = push(head->right->right, 50);` `    ``head->right->right = push(head->right->right, 22);` `    ``head->right->right = push(head->right->right, 19);`   `    ``head->right->right->right = push(head->right->right->right, 45);` `    ``head->right->right->right = push(head->right->right->right, 40);` `    ``head->right->right->right = push(head->right->right->right, 35);` `    ``head->right->right->right = push(head->right->right->right, 20);`   `    ``// Flatten the list` `    ``head = flatten(head);`   `    ``printList();` `    ``return` `0;` `}`   `// This code is contributed by rajsanghavi9.`

## Java

 `// Java program for flattening a Linked List` `class` `LinkedList` `{` `    ``Node head;  ``// head of list`   `    ``/* Linked list Node*/` `    ``class` `Node` `    ``{` `        ``int` `data;` `        ``Node right, down;` `        ``Node(``int` `data)` `        ``{` `            ``this``.data = data;` `            ``right = ``null``;` `            ``down = ``null``;` `        ``}` `    ``}`   `    ``// An utility function to merge two sorted linked lists` `    ``Node merge(Node a, Node b)` `    ``{` `        ``// if first linked list is empty then second` `        ``// is the answer` `        ``if` `(a == ``null``)     ``return` `b;`   `        ``// if second linked list is empty then first` `        ``// is the result` `        ``if` `(b == ``null``)      ``return` `a;`   `        ``// compare the data members of the two linked lists` `        ``// and put the larger one in the result` `        ``Node result;`   `        ``if` `(a.data < b.data)` `        ``{` `            ``result = a;` `            ``result.down =  merge(a.down, b);` `        ``}`   `        ``else` `        ``{` `            ``result = b;` `            ``result.down = merge(a, b.down);` `        ``}`   `        ``result.right = ``null``; ` `        ``return` `result;` `    ``}`   `    ``Node flatten(Node root)` `    ``{` `        ``// Base Cases` `        ``if` `(root == ``null` `|| root.right == ``null``)` `            ``return` `root;`   `        ``// recur for list on right` `        ``root.right = flatten(root.right);`   `        ``// now merge` `        ``root = merge(root, root.right);`   `        ``// return the root` `        ``// it will be in turn merged with its left` `        ``return` `root;` `    ``}`   `    ``/* Utility function to insert a node at beginning of the` `       ``linked list */` `    ``Node push(Node head_ref, ``int` `data)` `    ``{` `        ``/* 1 & 2: Allocate the Node &` `                  ``Put in the data*/` `        ``Node new_node = ``new` `Node(data);`   `        ``/* 3. 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;` `    ``}`   `    ``void` `printList()` `    ``{` `        ``Node temp = head;` `        ``while` `(temp != ``null``)` `        ``{` `            ``System.out.print(temp.data + ``" "``);` `            ``temp = temp.down;` `        ``}` `        ``System.out.println();` `    ``}`   `    ``/* Driver program to test above functions */` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``LinkedList L = ``new` `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 Rajat Mishra */`

## 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 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 program to test above functions ` `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`

## C#

 `// C# program for flattening a Linked List` `using` `System;` `public` `class` `List {` `    ``Node head; ``// head of list`   `    ``/* Linked list Node */` `    ``public`     ` ``class` `Node {` `        ``public`     ` ``int` `data;` `        ``public`     ` ``Node right, down;`   `        ``public`     ` ``Node(``int` `data) {` `            ``this``.data = data;` `            ``right = ``null``;` `            ``down = ``null``;` `        ``}` `    ``}`   `    ``// An utility function to merge two sorted linked lists` `    ``Node merge(Node a, Node b) {` `        ``// if first linked list is empty then second` `        ``// is the answer` `        ``if` `(a == ``null``)` `            ``return` `b;`   `        ``// if second linked list is empty then first` `        ``// is the result` `        ``if` `(b == ``null``)` `            ``return` `a;`   `        ``// compare the data members of the two linked lists` `        ``// and put the larger one in the result` `        ``Node result;`   `        ``if` `(a.data < b.data) {` `            ``result = a;` `            ``result.down = merge(a.down, b);` `        ``}`   `        ``else` `{` `            ``result = b;` `            ``result.down = merge(a, b.down);` `        ``}`   `        ``result.right = ``null``;` `        ``return` `result;` `    ``}`   `    ``Node flatten(Node root) {` `        ``// Base Cases` `        ``if` `(root == ``null` `|| root.right == ``null``)` `            ``return` `root;`   `        ``// recur for list on right` `        ``root.right = flatten(root.right);`   `        ``// now merge` `        ``root = merge(root, root.right);`   `        ``// return the root` `        ``// it will be in turn merged with its left` `        ``return` `root;` `    ``}`   `    ``/*` `     ``* Utility function to insert a node at beginning of the linked list` `     ``*/` `    ``Node Push(Node head_ref, ``int` `data) {` `        ``/*` `         ``* 1 & 2: Allocate the Node & Put in the data` `         ``*/` `        ``Node new_node = ``new` `Node(data);`   `        ``/* 3. 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;` `    ``}`   `    ``void` `printList() {` `        ``Node temp = head;` `        ``while` `(temp != ``null``) {` `            ``Console.Write(temp.data + ``" "``);` `            ``temp = temp.down;` `        ``}` `        ``Console.WriteLine();` `    ``}`   `    ``/* Driver program to test above functions */` `    ``public` `static` `void` `Main(String []args) {` `        ``List L = ``new` `List();`   `        ``/*` `         ``* 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 umadevi9616`

## Javascript

 ``

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).

Explanation: As we are merging 2 lists at a time,

• After adding first 2 lists, time taken will be O(M+M) = O(2M).
• Then we will merge another list to above merged list -> time = O(2M + M) = O(3M).
• Then we will merge another list -> time = O(3M + M).
• We will keep merging lists to previously merged lists until all lists are merged.
• Total time taken will be O(2M + 3M + 4M + …. N*M) = (2 + 3 + 4 + … + N)*M
• Using arithmetic sum formula: time = O((N*N + N – 2)*M/2)
• Above expression is roughly equal to O(N*N*M) for large value of N

Space Complexity: O(N*M) – because of the recursion. The recursive functions will use recursive stack of size equivalent to total number of elements in the lists, which is N*M.

#### Method 2: Using Priority Queues

The idea is, to build min-heap and  use Extract-min property  to get minimum element from priority queue.

## C++

 `#include `   `struct` `Node {` `    ``int` `data;` `    ``struct` `Node* next;` `    ``struct` `Node* bottom;`   `    ``Node(``int` `x)` `    ``{` `        ``data = x;` `        ``next = NULL;` `        ``bottom = NULL;` `    ``}` `};`   `using` `namespace` `std;`   `void` `flatten(Node* root);`   `int` `main(``void``)` `{` `  ``//this code builds the flattened linked list` `  ``//of first picture in this article ;` `  ``Node* head=``new` `Node(5);` `  ``auto` `temp=head;` `  ``auto` `bt=head;` `  ``bt->bottom=``new` `Node(7);` `  ``bt->bottom->bottom=``new` `Node(8);` `  ``bt->bottom->bottom->bottom=``new` `Node(30);` `  ``temp->next=``new` `Node(10);` `  `  `  ``temp=temp->next;` `  ``bt=temp;` `  ``bt->bottom=``new` `Node(20);` `  ``temp->next=``new` `Node(19);` `  ``temp=temp->next;` `  ``bt=temp;` `  ``bt->bottom=``new` `Node(22);` `  ``bt->bottom->bottom=``new` `Node(50);` `  ``temp->next=``new` `Node(28);` `  ``temp=temp->next;` `  ``bt=temp;` `  ``bt->bottom=``new` `Node(35);` `  ``bt->bottom->bottom=``new` `Node(40);` `  ``bt->bottom->bottom->bottom=``new` `Node(45);` `  `  `  ``flatten(head);` `  ``cout << endl;` `  ``return` `0;` `}`       `struct` `mycomp {` `    ``bool` `operator()(Node* a, Node* b)` `    ``{` `        ``return` `a->data > b->data;` `    ``}` `};` `void` `flatten(Node* root)` `{` `    ``priority_queue, mycomp> p;` `  ``//pushing main link nodes into priority_queue.` `    ``while` `(root!=NULL) {` `        ``p.push(root);` `        ``root = root->next;` `    ``}` `  `  `    ``while` `(!p.empty()) {` `      ``//extracting min` `        ``auto` `k = p.top();` `        ``p.pop();` `      ``//printing  least element ` `        ``cout << k->data << ``" "``;` `        ``if` `(k->bottom)` `            ``p.push(k->bottom);` `    ``}` `   `  `}` `//this code is contributed by user_990i`

Output

`5 7 8 10 19 20 22 28 30 35 40 45 50 `

Time Complexity: O(N*M*log(N)) – where N is the no of nodes in main linked list (reachable using next pointer) and M is the no of node in a single sub linked list (reachable using bottom pointer).

Explanation:

1. Push all the nodes that are reachable using next pointer into priority queue till we encounter NULL

2.while(priority_queue is not empty)

(i) Extract Node which contains minimum-value(k) from priority_queue and print k->data.

(ii) If  k->bottom!=NULL ,insert k->bottom into priority_queue. Else our priority_queue size will decrease.

Space Complexity: O(N)-where N is the no of nodes in main linked list (reachable using next pointer).

NOTE: In above explanation, k means the Node which contains minimum-element.

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