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Practice questions for Linked List and Recursion
• Difficulty Level : Basic
• Last Updated : 03 Jun, 2021

Assume the structure of a Linked List node is as follows.

## C++

 `struct` `Node``{``  ``int` `data;``  ``struct` `Node *next;``};` `// This code is contributed by SHUBHAMSINGH10`

## C

 `struct` `Node``{``  ``int` `data;``  ``struct` `Node *next;``};`

## Java

 `static` `class` `Node``{``    ``int` `data;``    ``Node next;``};` `// This code is contributed by shubhamsingh10`

## Python3

 `class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None`

## C#

 `public` `class` `Node``{ ``    ``public` `int` `data;``    ``public` `Node next;``};` `// This code is contributed by pratham_76`

## Javascript

 ``

Explain the functionality of the following C functions.

1. What does the following function do for a given Linked List?

## C++14

 `void` `fun1(``struct` `Node* head)``{``  ``if` `(head == NULL)``      ``return``;` `  ``fun1(head->next);``  ``cout << head->data << ``" "``;``}` `// This code is contributed by shubhamsingh10`

## C

 `void` `fun1(``struct` `Node* head)``{``  ``if``(head == NULL)``    ``return``;`` ` `  ``fun1(head->next);``  ``printf``(``"%d  "``, head->data);``}`

## Java

 `static` `void` `fun1(Node head)``{``    ``if` `(head == ``null``)``    ``{``        ``return``;``    ``}` `    ``fun1(head.next);``    ``System.out.print(head.data + ``" "``);``}` `// This code is contributed by shubhamsingh10`

## Python

 `def` `fun1(head):``    ``if``(head ``=``=` `None``):``        ``return``    ``fun1(head.``next``)``    ``print``(head.data, end ``=` `" "``)` `# This code is contributed by shubhamsingh10`

## C#

 `static` `void` `fun1(Node head)``{``    ``if` `(head == ``null``)``    ``{``        ``return``;``    ``}` `    ``fun1(head.next);``    ``Console.Write(head.data + ``" "``);``}` `// This code is contributed by shubhamsingh10`

## Javascript

 ``

fun1() prints the given Linked List in the reverse way. For Linked List 1->2->3->4->5, fun1() prints 5->4->3->2->1.

2. What does the following function do for a given Linked List?

## C++

 `void` `fun2(``struct` `Node* head)``{``    ``if``(head == NULL)``        ``return``;``    ``cout << head->data << ``" "``;``    ` `    ``if``(head->next != NULL )``        ``fun2(head->next->next);``    ``cout << head->data << ``" "``;``}` `// This code is contributed by shubhamsingh10`

## C

 `void` `fun2(``struct` `Node* head)``{``  ``if``(head == NULL)``    ``return``;``  ``printf``(``"%d  "``, head->data);` `  ``if``(head->next != NULL )``    ``fun2(head->next->next);``  ``printf``(``"%d  "``, head->data);  ``}`

## Java

 `static` `void` `fun2(Node head)``{``    ``if` `(head == ``null``)``    ``{``        ``return``;``    ``}``    ``System.out.print(head.data + ``" "``);` `    ``if` `(head.next != ``null``)``    ``{``        ``fun2(head.next.next);``    ``}``    ``System.out.print(head.data + ``" "``);``}` `// This code is contributed by shubhamsingh10`

## Python3

 `def` `fun2(head):``     ` `    ``if``(head ``=``=` `None``):``        ``return``    ``print``(head.data, end ``=` `" "``)``     ` `    ``if``(head.``next` `!``=` `None` `):``        ``fun2(head.``next``.``next``)``    ``print``(head.data, end ``=` `" "``)` `    ``# This code is contributed by divyesh072019`

## C#

 `static` `void` `fun2(Node head)``{``    ``if` `(head == ``null``)``    ``{``        ``return``;``    ``}``    ``Console.Write(head.data + ``" "``);` `    ``if` `(head.next != ``null``)``    ``{``        ``fun2(head.next.next);``    ``}``    ``Console.Write(head.data + ``" "``);``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

fun2() prints alternate nodes of the given Linked List, first from head to end, and then from end to head. If Linked List has even number of nodes, then fun2() skips the last node. For Linked List 1->2->3->4->5, fun2() prints 1 3 5 5 3 1. For Linked List 1->2->3->4->5->6, fun2() prints 1 3 5 5 3 1.

Below is a complete running program to test the above functions.

## C++

 `#include ``using` `namespace` `std;` `/* A linked list node */``class` `Node``{``    ``public``:``    ``int` `data;``    ``Node *next;``};`  `/* Prints a linked list in reverse manner */``void` `fun1(Node* head)``{``    ``if``(head == NULL)``        ``return``;``    ` `    ``fun1(head->next);``    ``cout << head->data << ``" "``;``}` `/* prints alternate nodes of a Linked List, first``from head to end, and then from end to head. */``void` `fun2(Node* start)``{``    ``if``(start == NULL)``        ``return``;``    ``cout<data<<``" "``;``    ` `    ``if``(start->next != NULL )``        ``fun2(start->next->next);``    ``cout << start->data << ``" "``;``}` `/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */``/* Given a reference (pointer to pointer) to the head``of a list and an int, push a new node on the front``of the list. */``void` `push(Node** head_ref, ``int` `new_data)``{``    ``/* allocate node */``    ``Node* new_node = ``new` `Node();``    ` `    ``/* put in the data */``    ``new_node->data = new_data;``    ` `    ``/* link the old list off the new node */``    ``new_node->next = (*head_ref);``    ` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Driver code */``int` `main()``{``    ``/* Start with the empty list */``    ``Node* head = NULL;``    ` `    ``/* Using push() to construct below list``        ``1->2->3->4->5 */``    ``push(&head, 5);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 1);``    ` `    ``cout<<``"Output of fun1() for list 1->2->3->4->5 \n"``;``    ``fun1(head);``    ` `    ``cout<<``"\nOutput of fun2() for list 1->2->3->4->5 \n"``;``    ``fun2(head);` `    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `#include``#include` `/* A linked list node */``struct` `Node``{``  ``int` `data;``  ``struct` `Node *next;``};`  `/* Prints a linked list in reverse manner */``void` `fun1(``struct` `Node* head)``{``  ``if``(head == NULL)``    ``return``;` `  ``fun1(head->next);``  ``printf``(``"%d  "``, head->data);``}` `/* prints alternate nodes of a Linked List, first``  ``from head to end, and then from end to head. */``void` `fun2(``struct` `Node* start)``{``  ``if``(start == NULL)``    ``return``;``  ``printf``(``"%d  "``, start->data);` `  ``if``(start->next != NULL )``    ``fun2(start->next->next);``  ``printf``(``"%d  "``, start->data);``}` `/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */``/* Given a reference (pointer to pointer) to the head``  ``of a list and an int, push a new node on the front``  ``of the list. */``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{``  ``/* allocate node */``  ``struct` `Node* new_node =``          ``(``struct` `Node*) ``malloc``(``sizeof``(``struct` `Node));`` ` `  ``/* put in the data  */``  ``new_node->data  = new_data;`` ` `  ``/* link the old list off the new node */``  ``new_node->next = (*head_ref);`` ` `  ``/* move the head to point to the new node */``  ``(*head_ref)    = new_node;``}`` ` `/* Driver program to test above functions */``int` `main()``{``  ``/* Start with the empty list */``  ``struct` `Node* head = NULL;` `  ``/* Using push() to construct below list``    ``1->2->3->4->5  */``  ``push(&head, 5);``  ``push(&head, 4);``  ``push(&head, 3);``  ``push(&head, 2);``  ``push(&head, 1);  `` ` `  ``printf``(``"Output of fun1() for list 1->2->3->4->5 \n"``);``  ``fun1(head);` `  ``printf``(``"\nOutput of fun2() for list 1->2->3->4->5 \n"``);``  ``fun2(head);``        ` `  ``getchar``();``  ``return` `0;``}`

## Java

 `// Java code implementation for above approach``class` `GFG``{` `    ``/* A linked list node */``    ``static` `class` `Node``    ``{``        ``int` `data;``        ``Node next;``    ``};` `    ``/* Prints a linked list in reverse manner */``    ``static` `void` `fun1(Node head)``    ``{``        ``if` `(head == ``null``)``        ``{``            ``return``;``        ``}` `        ``fun1(head.next);``        ``System.out.print(head.data + ``" "``);``    ``}` `    ``/* prints alternate nodes of a Linked List, first``    ``from head to end, and then from end to head. */``    ``static` `void` `fun2(Node start)``    ``{``        ``if` `(start == ``null``)``        ``{``            ``return``;``        ``}``        ``System.out.print(start.data + ``" "``);` `        ``if` `(start.next != ``null``)``        ``{``            ``fun2(start.next.next);``        ``}``        ``System.out.print(start.data + ``" "``);``    ``}` `    ``/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */``    ``/* Given a reference (pointer to pointer) to the head``    ``of a list and an int, push a new node on the front``    ``of the list. */``    ``static` `Node push(Node head_ref, ``int` `new_data)``    ``{``        ``/* allocate node */``        ``Node new_node = ``new` `Node();` `        ``/* put in the data */``        ``new_node.data = new_data;` `        ``/* link the old list off the new node */``        ``new_node.next = (head_ref);` `        ``/* move the head to point to the new node */``        ``(head_ref) = new_node;``        ``return` `head_ref;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``/* Start with the empty list */``        ``Node head = ``null``;` `        ``/* Using push() to conbelow list``        ``1->2->3->4->5 */``        ``head = push(head, ``5``);``        ``head = push(head, ``4``);``        ``head = push(head, ``3``);``        ``head = push(head, ``2``);``        ``head = push(head, ``1``);` `        ``System.out.print(``"Output of fun1() for "` `+``                         ``"list 1->2->3->4->5 \n"``);``        ``fun1(head);` `        ``System.out.print(``"\nOutput of fun2() for "` `+``                           ``"list 1->2->3->4->5 \n"``);``        ``fun2(head);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `''' A linked list node '''``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `''' Prints a linked list in reverse manner '''``def` `fun1(head):``    ``if``(head ``=``=` `None``):``        ``return``    ``fun1(head.``next``)``    ``print``(head.data, end ``=` `" "``)` `''' prints alternate nodes of a Linked List, first``from head to end, and then from end to head. '''``def` `fun2(start):``    ` `    ``if``(start ``=``=` `None``):``        ``return``    ``print``(start.data, end ``=` `" "``)``    ` `    ``if``(start.``next` `!``=` `None` `):``        ``fun2(start.``next``.``next``)``    ``print``(start.data, end ``=` `" "``)`  `''' UTILITY FUNCTIONS TO TEST fun1() and fun2() '''``''' Given a reference (pointer to pointer) to the head``of a list and an int, push a new node on the front``of the list. '''``def` `push( head, new_data):``    ` `    ``''' put in the data '''``    ``new_node ``=` `Node(new_data)``    ` `    ``''' link the old list off the new node '''``    ``new_node.``next` `=` `head``    ` `    ``''' move the head to poto the new node '''``    ``head ``=` `new_node``    ``return` `head` `''' Driver code '''``''' Start with the empty list '''``head ``=` `None` `''' Using push() to construct below list``    ``1.2.3.4.5 '''``head ``=` `Node(``5``)``head ``=` `push(head, ``4``)``head ``=` `push(head, ``3``)``head ``=` `push(head, ``2``)``head ``=` `push(head, ``1``)` `print``(``"Output of fun1() for list 1->2->3->4->5"``)``fun1(head)` `print``(``"\nOutput of fun2() for list 1->2->3->4->5"``)``fun2(head)` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# code implementation for above approach``using` `System;` `class` `GFG``{` `    ``/* A linked list node */``    ``public` `class` `Node``    ``{``        ``public` `int` `data;``        ``public` `Node next;``    ``};` `    ``/* Prints a linked list in reverse manner */``    ``static` `void` `fun1(Node head)``    ``{``        ``if` `(head == ``null``)``        ``{``            ``return``;``        ``}` `        ``fun1(head.next);``        ``Console.Write(head.data + ``" "``);``    ``}` `    ``/* prints alternate nodes of a Linked List, first``    ``from head to end, and then from end to head. */``    ``static` `void` `fun2(Node start)``    ``{``        ``if` `(start == ``null``)``        ``{``            ``return``;``        ``}``        ``Console.Write(start.data + ``" "``);` `        ``if` `(start.next != ``null``)``        ``{``            ``fun2(start.next.next);``        ``}``        ``Console.Write(start.data + ``" "``);``    ``}` `    ``/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */``    ``/* Given a reference (pointer to pointer) to the head``    ``of a list and an int,.Push a new node on the front``    ``of the list. */``    ``static` `Node Push(Node head_ref, ``int` `new_data)``    ``{``        ``/* allocate node */``        ``Node new_node = ``new` `Node();` `        ``/* put in the data */``        ``new_node.data = new_data;` `        ``/* link the old list off the new node */``        ``new_node.next = (head_ref);` `        ``/* move the head to point to the new node */``        ``(head_ref) = new_node;``        ``return` `head_ref;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``/* Start with the empty list */``        ``Node head = ``null``;` `        ``/* Using.Push() to conbelow list``        ``1->2->3->4->5 */``        ``head = Push(head, 5);``        ``head = Push(head, 4);``        ``head = Push(head, 3);``        ``head = Push(head, 2);``        ``head = Push(head, 1);` `        ``Console.Write(``"Output of fun1() for "` `+``                        ``"list 1->2->3->4->5 \n"``);``        ``fun1(head);` `        ``Console.Write(``"\nOutput of fun2() for "` `+``                        ``"list 1->2->3->4->5 \n"``);``        ``fun2(head);``    ``}``}` `// This code is contributed by Rajput-Ji`

Output:

``` Output of fun1() for list 1->2->3->4->5
5 4 3 2 1
Output of fun2() for list 1->2->3->4->5
1 3 5 5 3 1```

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