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:
// 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 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 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# 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
<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)
