Delete nodes which have a greater value on right side using recursion

Given a singly linked list, remove all the nodes which have a greater value on the right side. 
Examples:
a) The list 12->15->10->11->5->6->2->3->NULL should be changed to 15->11->6->3->NULL. Note that 12, 10, 5 and 2 have been deleted because there is a greater value on the right side. 
When we examine 12, we see that after 12 there is one node with a value greater than 12 (i.e. 15), so we delete 12. 
When we examine 15, we find no node after 15 that has a value greater than 15 so we keep this node. 
When we go like this, we get 15->6->3
b) The list 10->20->30->40->50->60->NULL should be changed to 60->NULL. Note that 10, 20, 30, 40 and 50 have been deleted because they all have a greater value on the right side.
c) The list 60->50->40->30->20->10->NULL should not be changed.

Approach: We have already solved this problem by using 2 loops and reversing linked list in the post Delete nodes which have a greater value on right side
Here we will discuss the solution without reversing the list. We will use recursion to solve this problem in which the base case would be when the head is pointing to NULL. Else we would be recursively calling function for the next node and updating max value if currentNode->data > currentMax. In this way, the whole list would be updated.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

/* structure of a linked list node */
struct Node {
    int data;
    struct Node* next;
};

/*Utility function to find maximum value*/
int maxVal(int a, int b)
{
    if (a > b)
        return a;
    return b;
}

/* Function to delete nodes which have
a node with greater value node
on left side */
struct Node* delNodes(struct Node* head, int* max)
{

    // Base case
    if (head == NULL) {
        return head;
    }

    head->next = delNodes(head->next, max);
    if (head->data < *max) {
        return head->next;
    }
    *max = maxVal(head->data, *max);
    return head;
}

/* Utility function to insert a node at the beginning */
void push(struct Node** head, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = *head;
    *head = new_node;
}

/* Utility function to print a linked list */
void printList(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}

/* Driver program to test above functions */
int main()
{
    struct Node* head = NULL;

    /* Create following linked list
    12->15->10->11->5->6->2->3 */
    push(&head, 3);
    push(&head, 2);
    push(&head, 6);
    push(&head, 5);
    push(&head, 11);
    push(&head, 10);
    push(&head, 15);
    push(&head, 12);

    cout << "Given Linked List" << endl;
    printList(head);
    int max = INT_MIN;
    head = delNodes(head, &max);

    cout << "Modified Linked List" << endl;
    printList(head);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

/* structure of a linked list node */
static class Node
{
    int data;
    Node next;
};
static Node head;
static int max;

/*Utility function to find maximum value*/
static int maxVal(int a, int b)
{
    if (a > b)
        return a;
    return b;
}

/* Function to delete nodes which have
a node with greater value node
on left side */
static Node delNodes(Node head)
{

    // Base case
    if (head == null)
    {
        return head;
    }

    head.next = delNodes(head.next);
    if (head.data < max)
    {
        return head.next;
    }
    max = maxVal(head.data, max);
    return head;
}

/* Utility function to insert a node
at the beginning */
static void 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;
        head = head_ref;
}

/* Utility function to print a linked list */
static void printList(Node head)
{
    while (head != null)
    {
        System.out.print(head.data + " ");
        head = head.next;
    }
    System.out.println();
}

// Driver Code
public static void main(String[] args)
{
    head = null;

    /* Create following linked list
    12.15.10.11.5.6.2.3 */
    push(head, 3);
    push(head, 2);
    push(head, 6);
    push(head, 5);
    push(head, 11);
    push(head, 10);
    push(head, 15);
    push(head, 12);

    System.out.println("Given Linked List");
    printList(head);
    max = Integer.MIN_VALUE;
    head = delNodes(head);

    System.out.println("Modified Linked List");
    printList(head);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to reverse a linked
# list using a stack

# Link list node
class Node:

    def __init__(self, data, next):
        self.data = data
        self.next = next

class LinkedList:

    def __init__(self):
        self.head = None

    # Function to push a new Node in
    # the linked list
    def push(self, new_data):

        new_node = Node(new_data, self.head)
        self.head = new_node

    # Function to delete nodes which have a node
    # with greater value node on left side
    def delNodes(self, head):

        # Base case
        if head == None:
            return head

        global Max

        head.next = self.delNodes(head.next)
        if head.data < Max:
            return head.next

        Max = max(head.data, Max)
        return head

    # Function to print the Linked list
    def printList(self):

        curr = self.head
        while curr:
            print(curr.data, end = " ")
            curr = curr.next
        print()

# Driver Code
if __name__ == "__main__":

    # Start with the empty list
    linkedList = LinkedList()

    # Create following linked list
    # 12->15->10->11->5->6->2->3
    linkedList.push(3)
    linkedList.push(2)
    linkedList.push(6)
    linkedList.push(5)
    linkedList.push(11)
    linkedList.push(10)
    linkedList.push(15)
    linkedList.push(12)

    print("Given Linked List")
    linkedList.printList()
    Max = float('-inf')
    linkedList.head = linkedList.delNodes(linkedList.head)

    print("Modified Linked List")

    linkedList.printList()

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

/* structure of a linked list node */
public class Node
{
    public int data;
    public Node next;
};
static Node head;
static int max;

/*Utility function to find maximum value*/
static int maxVal(int a, int b)
{
    if (a > b)
        return a;
    return b;
}

/* Function to delete nodes which have
a node with greater value node
on left side */
static Node delNodes(Node head)
{

    // Base case
    if (head == null)
    {
        return head;
    }

    head.next = delNodes(head.next);
    if (head.data < max)
    {
        return head.next;
    }
    max = maxVal(head.data, max);
    return head;
}

/* Utility function to insert a node
at the beginning */
static void 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;
        head = head_ref;
}

/* Utility function to print a linked list */
static void printList(Node head)
{
    while (head != null)
    {
        Console.Write(head.data + " ");
        head = head.next;
    }
    Console.WriteLine();
}

// Driver Code
public static void Main(String[] args)
{
    head = null;

    /* Create following linked list
    12.15.10.11.5.6.2.3 */
    push(head, 3);
    push(head, 2);
    push(head, 6);
    push(head, 5);
    push(head, 11);
    push(head, 10);
    push(head, 15);
    push(head, 12);

    Console.WriteLine("Given Linked List");
    printList(head);
    max = int.MinValue;
    head = delNodes(head);

    Console.WriteLine("Modified Linked List");
    printList(head);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

      // JavaScript implementation of the approach
      /* structure of a linked list node */
      class Node {
        constructor() {
          this.data = 0;
          this.next = null;
        }
      }
      var head;
      var max;

      /*Utility function to find maximum value*/
      function maxVal(a, b) {
        if (a > b) return a;
        return b;
      }

      /* Function to delete nodes which have
         a node with greater value node
         on left side */
      function delNodes(head) {
        // Base case
        if (head == null) {
          return head;
        }

        head.next = delNodes(head.next);
        if (head.data < max) {
          return head.next;
        }
        max = maxVal(head.data, max);
        return head;
      }

      /* Utility function to insert a 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;
        head = head_ref;
      }

      /* Utility function to print a linked list */
      function printList(head) {
        while (head != null) {
          document.write(head.data + " ");
          head = head.next;
        }
        document.write("<br>");
      }

      // Driver Code
      head = null;

      /* Create following linked list
      12.15.10.11.5.6.2.3 */
      push(head, 3);
      push(head, 2);
      push(head, 6);
      push(head, 5);
      push(head, 11);
      push(head, 10);
      push(head, 15);
      push(head, 12);

      document.write("Given Linked List <br>");
      printList(head);
      max = -2147483648;
      head = delNodes(head);

      document.write("Modified Linked List <br>");
      printList(head);

</script>

Output:

Given Linked List
12 15 10 11 5 6 2 3 
Modified Linked List
15 11 6 3

Time Complexity: O(N)

Auxiliary Space: O(1)