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Python Program For Merge Sort Of Linked Lists

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  • Last Updated : 18 May, 2022
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Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible. 

sorting image

Let the head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so the head node has to be changed if the data at the original head is not the smallest value in the linked list. 

1) If the head is NULL or there is only one element in the Linked List 
    then return.
2) Else divide the linked list into two halves.  
      FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
4) Merge the sorted a and b (using SortedMerge() discussed here) 
   and update the head pointer using headRef.
     *headRef = SortedMerge(a, b);


# Python3 program to merge sort of linked list
# create Node using class Node.
class Node:
    def __init__(self, data): = data = None
class LinkedList:
    def __init__(self):
        self.head = None
    # push new value to linked list
    # using append method
    def append(self, new_value):
        # Allocate new node
        new_node = Node(new_value)
        # if head is None, initialize it to new node
        if self.head is None:
            self.head = new_node
        curr_node = self.head
        while is not None:
            curr_node =
        # Append the new node at the end
        # of the linked list = new_node
    def sortedMerge(self, a, b):
        result = None
        # Base cases
        if a == None:
            return b
        if b == None:
            return a
        # pick either a or b and recur..
        if <=
            result = a
   = self.sortedMerge(, b)
            result = b
   = self.sortedMerge(a,
        return result
    def mergeSort(self, h):
        # Base case if head is None
        if h == None or == None:
            return h
        # get the middle of the list 
        middle = self.getMiddle(h)
        nexttomiddle =
        # set the next of middle node to None = None
        # Apply mergeSort on left list 
        left = self.mergeSort(h)
        # Apply mergeSort on right list
        right = self.mergeSort(nexttomiddle)
        # Merge the left and right lists 
        sortedlist = self.sortedMerge(left, right)
        return sortedlist
    # Utility function to get the middle 
    # of the linked list 
    def getMiddle(self, head):
        if (head == None):
            return head
        slow = head
        fast = head
        while ( != None and 
      != None):
            slow =
            fast =
        return slow
# Utility function to print the linked list 
def printList(head):
    if head is None:
        print(' ')
    curr_node = head
    while curr_node:
        print(, end = " ")
        curr_node =
    print(' ')
# Driver Code
if __name__ == '__main__':
    li = LinkedList()
    # Let us create a unsorted linked list
    # to test the functions created. 
    # The list shall be a: 2->3->20->5->10->15 
    # Apply merge Sort 
    li.head = li.mergeSort(li.head)
    print ("Sorted Linked List is:")
# This code is contributed by Vikas Chitturi


Sorted Linked List is: 
2 3 5 10 15 20


Time Complexity: O(n*log n)

Space Complexity: O(n*log n)

Approach 2: This approach is simpler and uses log n space.


  1. If the size of the linked list is 1 then return the head
  2. Find mid using The Tortoise and The Hare Approach
  3. Store the next of mid in head2 i.e. the right sub-linked list.
  4. Now Make the next midpoint null.
  5. Recursively call mergeSort() on both left and right sub-linked list and store the new head of the left and right linked list.
  6. Call merge() given the arguments new heads of left and right sub-linked lists and store the final head returned after merging.
  7. Return the final head of the merged linkedlist.

merge(head1, head2):

  1. Take a pointer say merged to store the merged list in it and store a dummy node in it.
  2. Take a pointer temp and assign merge to it.
  3. If the data of head1 is less than the data of head2, then, store head1 in next of temp & move head1 to the next of head1.
  4. Else store head2 in next of temp & move head2 to the next of head2.
  5. Move temp to the next of temp.
  6. Repeat steps 3, 4 & 5 until head1 is not equal to null and head2 is not equal to null.
  7. Now add any remaining nodes of the first or the second linked list to the merged linked list.
  8. Return the next of merged(that will ignore the dummy and return the head of the final merged linked list)


# Python program for the above approach
# Node Class
class Node:
    def __init__(self,key):
# Function to merge sort
def mergeSort(head):
    if ( == None):
        return head
    mid = findMid(head)
    head2 = = None
    newHead1 = mergeSort(head)
    newHead2 = mergeSort(head2)
    finalHead = merge(newHead1, newHead2)
    return finalHead
# Function to merge two linked lists
def merge(head1,head2):
    merged = Node(-1)
    temp = merged
    # While head1 is not null and head2
    # is not null
    while (head1 != None and head2 != None):
        if ( <
   = head1
            head1 =
   = head2
            head2 =
        temp =
    # While head1 is not null
    while (head1 != None): = head1
        head1 =
        temp =
    # While head2 is not null
    while (head2 != None): = head2
        head2 =
        temp =
# Find mid using The Tortoise and The Hare approach
def findMid(head):
    slow = head
    fast =
    while (fast != None and != None):
        slow =
        fast =
    return slow
# Function to print list
def printList(head):
    while (head != None):
        print(,end=" ")
# Driver Code
head = Node(7)
temp = head = Node(10);
temp =; = Node(5);
temp =; = Node(20);
temp =; = Node(3);
temp =; = Node(2);
temp =;
# Apply merge Sort
head = mergeSort(head);
Sorted Linked List is
# This code is contributed by avanitrachhadiya2155


Sorted Linked List is: 
2 3 5 7 10 20 

Time Complexity: O(n*log n)

Space Complexity: O(log n)

Please refer complete article on Merge Sort for Linked Lists for more details!

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