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Practice questions for Linked List and Recursion

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  • Difficulty Level : Easy
  • Last Updated : 24 Jun, 2022

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




<script>
  
class Node
{
    constructor(item)
    {
        this.data = item;
        this.next = null;
    }
}
   
// This code contributed by shubhamsingh10
</script>

Complete Interview Preparation - GFG

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




<script>
  
// Javascript Implementation
function fun1( head)
{
  if (head == null)
      return;
  
  fun1(head.next);
  document.write(head.data);
}
  
// This code is contributed by shubhamsingh10
</script>

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




<script>
  
// Javascript Implementation
function fun2( head)
{
  if (head == null)
      return;
  document.write(head.data);
    
  if (head.next != null)
      fun2(head.next.next);
  document.write(head.data);
}
  
// This code is contributed by shubhamsingh10
  
</script>

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 <bits/stdc++.h>
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<<start->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<stdio.h>
#include<stdlib.h>
  
/* 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 construct below 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 point to 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 construct below 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

Javascript




<script>
    // Javascript code implementation for above approach
      
    /* A linked list node */
    class Node
    {
        constructor(data) {
           this.next = null;
           this.data = data;
        }
    }
      
    /* Prints a linked list in reverse manner */
    function fun1(head)
    {
        if (head == null)
        {
            return;
        }
   
        fun1(head.next);
        document.write(head.data + " ");
    }
   
    /* prints alternate nodes of a Linked List, first
    from head to end, and then from end to head. */
    function fun2(start)
    {
        if (start == null)
        {
            return;
        }
        document.write(start.data + " ");
   
        if (start.next != null)
        {
            fun2(start.next.next);
        }
        document.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. */
    function Push(head_ref, new_data)
    {
        /* allocate node */
        /* put in the data */
        let new_node = new Node(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;
    }
      
    /* Start with the empty list */
    let head = null;
  
    /* Using.Push() to construct below 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);
  
    document.write("Output of fun1() for " +
                  "list 1->2->3->4->5 " + "</br>");
    fun1(head);
      
    document.write("</br>");
    document.write("Output of fun2() for " +
                  "list 1->2->3->4->5 " + "</br>");
    fun2(head);
     
   // This code is contributed by mukesh07.
</script>

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

Time complexity: O(n)

Auxiliary Space: O(1)

Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above.
 


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