Given a linked list, print alternate nodes of this linked list.
Examples :
Input : 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 Output : 1 -> 3 -> 5 -> 7 -> 9 Input : 10 -> 9 Output : 10
Recursive Approach :
- Initialize a static variable(say flag)
- If flag is odd print the node
- increase head and flag by 1, and recurse for next nodes.
Implementation:
C++
// CPP code to print alternate nodes // of a linked list using recursion #include <bits/stdc++.h> using namespace std;
// A linked list node struct Node {
int data;
struct Node* next;
}; // Inserting node at the beginning void push( struct Node** head_ref, int new_data)
{ struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
} // Function to print alternate nodes of linked list. // The boolean flag isOdd is used to find if the current // node is even or odd. void printAlternate( struct Node* node, bool isOdd= true )
{ if (node == NULL)
return ;
if (isOdd == true )
cout << node->data << " " ;
printAlternate(node->next, !isOdd);
} // Driver code int main()
{ // Start with the empty list
struct Node* head = NULL;
// construct below list
// 1->2->3->4->5->6->7->8->9->10
push(&head, 10);
push(&head, 9);
push(&head, 8);
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
printAlternate(head);
return 0;
} |
Java
// Java code to print alternate nodes // of a linked list using recursion class GFG
{ // A linked list node static class Node
{ int data;
Node next;
}; // Inserting node at the beginning static Node push( Node head_ref, int new_data)
{ Node new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
} // Function to print alternate nodes of linked list. // The boolean flag isOdd is used to find if the current // node is even or odd. static void printAlternate( Node node, boolean isOdd)
{ if (node == null )
return ;
if (isOdd == true )
System.out.print( node.data + " " );
printAlternate(node.next, !isOdd);
} // Driver code public static void main(String args[])
{ // Start with the empty list
Node head = null ;
// construct below list
// 1.2.3.4.5.6.7.8.9.10
head = push(head, 10 );
head = push(head, 9 );
head = push(head, 8 );
head = push(head, 7 );
head = push(head, 6 );
head = push(head, 5 );
head = push(head, 4 );
head = push(head, 3 );
head = push(head, 2 );
head = push(head, 1 );
printAlternate(head, true );
} } // This code is contributed by Arnab Kundu |
Python3
# Python3 code to print alternate nodes # of a linked list using recursion # A linked list node class Node:
def __init__( self , data):
self .data = data
self . next = None
# Inserting node at the beginning def push( head_ref, new_data):
new_node = Node(new_data);
new_node.data = new_data;
new_node. next = head_ref;
head_ref = new_node;
return head_ref;
# Function to print alternate nodes of # linked list. The boolean flag isOdd # is used to find if the current node # is even or odd. def printAlternate( node, isOdd):
if (node = = None ):
return ;
if (isOdd = = True ):
print ( node.data, end = " " );
if (isOdd = = True ):
isOdd = False ;
else :
isOdd = True ;
printAlternate(node. next , isOdd);
# Driver code # Start with the empty list head = None ;
# construct below list # 1->2->3->4->5->6->7->8->9->10 head = push(head, 10 );
head = push(head, 9 );
head = push(head, 8 );
head = push(head, 7 );
head = push(head, 6 );
head = push(head, 5 );
head = push(head, 4 );
head = push(head, 3 );
head = push(head, 2 );
head = push(head, 1 );
printAlternate(head, True );
# This code is contributed by 29AjayKumar |
C#
// C# code to print alternate nodes // of a linked list using recursion using System;
class GFG
{ // A linked list node public class Node
{ public int data;
public Node next;
}; // Inserting node at the beginning static Node push( Node head_ref, int new_data)
{ Node new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
} // Function to print alternate nodes of linked list. // The boolean flag isOdd is used to find if the current // node is even or odd. static void printAlternate( Node node, bool isOdd)
{ if (node == null )
return ;
if (isOdd == true )
Console.Write( node.data + " " );
printAlternate(node.next, !isOdd);
} // Driver code public static void Main(String []args)
{ // Start with the empty list
Node head = null ;
// construct below list
// 1.2.3.4.5.6.7.8.9.10
head = push(head, 10);
head = push(head, 9);
head = push(head, 8);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
printAlternate(head, true );
} } // This code has been contributed by 29AjayKumar |
Javascript
<script> // javascript code to print alternate nodes // of a linked list using recursion // A linked list node class Node { constructor(val) {
this .data = val;
this .next = null ;
}
} // Inserting node at the beginning
function push(head_ref , new_data) {
var new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
// Function to print alternate nodes of linked list.
// The boolean flag isOdd is used to find if the current
// node is even or odd.
function printAlternate(node, isOdd) {
if (node == null )
return ;
if (isOdd == true )
document.write(node.data + " " );
printAlternate(node.next, !isOdd);
}
// Driver code
// Start with the empty list
var head = null ;
// construct below list
// 1.2.3.4.5.6.7.8.9.10
head = push(head, 10);
head = push(head, 9);
head = push(head, 8);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
printAlternate(head, true );
// This code contributed by umadevi9616 </script> |
Output:
1 3 5 7 9
Time complexity: O(N) where N is no of nodes in linked list
Auxiliary space: O(1), If we consider recursive call stack then it would be O(n)