Python Program For Recursive Selection Sort For Singly Linked List - Swapping Node Links

Last Updated : 23 Jul, 2025

Given a singly linked list containing n nodes. The problem is to sort the list using the recursive selection sort technique. The approach should be such that it involves swapping node links instead of swapping node data.  

sorting image

Examples: 

Input: 10 -> 12 -> 8 -> 4 -> 6
Output: 4 -> 6 -> 8 -> 10 -> 12

In Selection Sort, we first find the minimum element, swap it with the beginning node and recur for the remaining list. Below is the recursive implementation of these steps for the linked list. 

recurSelectionSort(head)
     if head->next == NULL
         return head
     Initialize min = head
     Initialize beforeMin = NULL
     Initialize ptr = head
    
     while ptr->next != NULL 
         if min->data > ptr->next->data
         min = ptr->next
         beforeMin = ptr
     ptr = ptr->next    
    
     if min != head
         swapNodes(&head, head, min, beforeMin)
    
     head->next = recurSelectionSort(head->next)
     return head

swapNodes(head_ref, currX, currY, prevY)
     head_ref = currY
     prevY->next = currX

     Initialize temp = currY->next
     currY->next = currX->next
     currX->next  = temp

The swapNodes(head_ref, currX, currY, prevY) is based on the approach discussed here but it has been modified accordingly for the implementation of this post. 

Python 
# Python implementation of recursive
# selection sort for singly linked
# list | Swapping node links

# Linked List node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to swap nodes 'currX' and
# 'currY' in a linked list without
# swapping data
def swapNodes(head_ref, currX,
            currY, prevY):

    # Make 'currY' as new head
    head_ref = currY

    # Adjust links
    prevY.next = currX

    # Swap next pointers
    temp = currY.next
    currY.next = currX.next
    currX.next = temp
    return head_ref

# Function to sort the linked list using
# recursive selection sort technique
def recurSelectionSort(head):

    # If there is only a single node
    if (head.next == None):
        return head

    # 'min' - pointer to store the node
    # having minimum data value
    min = head

    # 'beforeMin' - pointer to store node
    # previous to 'min' node
    beforeMin = None
    ptr = head
    
    # Traverse the list till the last node
    while (ptr.next != None):
    
        # if true, then update 'min' and
        # 'beforeMin'
        if (ptr.next.data < min.data) :
        
            min = ptr.next
            beforeMin = ptr
            
        ptr = ptr.next
    
    # if 'min' and 'head' are not same,
    # swap the head node with the 'min' node
    if (min != head):
        head = swapNodes(head, head,
                        min, beforeMin)

    # Recursively sort the remaining list
    head.next =    recurSelectionSort(head.next)

    return head

# Function to sort the given linked list
def sort(head_ref):

    # If list is empty
    if ((head_ref) == None):
        return None

    # Sort the list using recursive
    # selection sort technique
    head_ref = recurSelectionSort(head_ref)
    return head_ref

# Function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):

    # Allocate node
    new_node = Node(0)

    # Put in the data
    new_node.data = new_data

    # Link the old list to the
    # new node
    new_node.next = (head_ref)

    # Move the head to point to the
    # new node
    (head_ref) = new_node
    return head_ref

# Function to print the linked list
def printList(head):

    while (head != None):    
        print(head.data , end = " ")
        head = head.next
    
# Driver code
head = None

# Create linked list 10.12.8.4.6
head = push(head, 6)
head = push(head, 4)
head = push(head, 8)
head = push(head, 12)
head = push(head, 10)

print("Linked list before sorting:")
printList(head)

# sort the linked list
head = sort(head)

print("\nLinked list after sorting:")
printList(head)

# This code is contributed by Arnab Kundu
# This code is modified by Susobhan Akhuli

Output
Linked list before sorting:
10 12 8 4 6 
Linked list after sorting:
4 6 8 10 12

Time Complexity: O(n2), since we are traversing the linked list using a while loop and inside it there is a recursive function is operating at O(n) time thus the complexity turns out to be quadratic
Auxiliary Space: O(n), since we are using recursion all elements in the linked List takes up the entire stack space

Please refer complete article on Recursive selection sort for singly linked list | Swapping node links for more details!

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