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Flatten a multilevel linked list using level order traversal

Last Updated : 03 Oct, 2024
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Given a linked list where in addition to the next pointer, each node has a child pointer, which may or may not point to a separate list. These child lists may have one or more children of their own to produce a multilevel linked list. Given the head of the first level of the list. The task is to flatten the list so that all the nodes appear in a single-level linked list. Flatten the list in a way that all nodes at the first level should come first, then nodes of the second level, and so on.

Examples:

Input:

2_5


Output: 1->4->6->2->5->7->3->8
Explanation: The multilevel linked list is flattened as it has no child pointers.

Approach:

To flatten a multilevel linked list, start from the top level and process each node sequentially. For each node, if it has a child node, append this child node to the end of the current list. Continue this process for every node, updating the end of the list accordingly, until all nodes are processed and the list is flattened.

Step by step implementation:

  • Initialize curr and tail pointer points to head node initially.
  • Start traversing from the first level and set the tail to the last node.
  • Start traversing from curr horizontally until curr is not NULL:
    • If curr->child is not equal to NULL, then append the child list to the end of the resultant list by using tail->next = curr->child. Traverse the child list horizontally , and set tail to the last node of the child list. Set curr->child = NULL to remove the link.
    • Move the curr pointer to the next node in the list.
  • Return the head node.

Following is the implementation of the above algorithm. 

C++ 
// C++ Program to flatten list with 
// next and child pointers 
#include <bits/stdc++.h> 
using namespace std; 

class Node { 
	public: 
	int data; 
	Node *next; 
	Node *child; 
	Node (int x) {
	    data = x;
	    next = nullptr;
      	child = nullptr;
	}
}; 


// function that flattens 
// a multilevel linked list
void flattenList(Node *head) {
    
	// Base case
	if (head == nullptr) 
	    return; 

	// Find tail node of first level
	Node *tail = head; 
	while (tail->next != nullptr) 
		tail = tail->next; 

	// One by one traverse through all nodes of first level 
	// linked list till we reach the tail node 
	Node *curr = head; 
	
	while (curr != nullptr) { 
	    
		// If current node has a child 
		if (curr->child) { 
		    
			// then append the child at the end of current list 
			tail->next = curr->child; 

			// and update the tail to new last node 
			Node* tmp = curr->child; 
			while (tmp->next) 
				tmp = tmp->next; 
			tail = tmp; 
			
			// Remove link between curr and child node
			curr->child = nullptr;
		} 

		// Change current node 
		curr = curr->next; 
	} 
} 

void printList(Node *head) { 
    Node* curr = head;
	while (curr != NULL) { 
		cout << curr->data << " "; 
		curr = curr->next; 
	} 
	cout<<endl; 
} 


int main() { 
  
  	//Linked List - 
  	// 1 -> 2 -> 3
	// |    |   
	// 4 -> 5   6
	// |
	// 7
  
	Node *head = new Node(1); 
	head->next = new Node(2);
	head->next->next = new Node(3);
	head->child = new Node(4);
	head->child->next = new Node(5);
	head->next->next->child = new Node(6); 
    head->child->child = new Node(7);

	flattenList(head); 
	printList(head); 
	return 0; 
}
C 
// C Program to flatten list with 
// next and child pointers
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
    struct Node* child;
};

// function that flattens 
// a multilevel linked list
void flattenList(struct Node *head) {
    
    // Base case
    if (head == NULL)
        return;

    // Find tail node of first level
    struct Node *tail = head;
    while (tail->next != NULL)
        tail = tail->next;

    // One by one traverse through all nodes of first level 
    // linked list till we reach the tail node
    struct Node *curr = head;

    while (curr != NULL) {
        
        // If current node has a child
        if (curr->child) {
            
            // then append the child at the end of current list
            tail->next = curr->child;

            // and update the tail to new last node
            struct Node *tmp = curr->child;
            while (tmp->next)
                tmp = tmp->next;
            tail = tmp;

            // Remove link between curr and child node
            curr->child = NULL;
        }

        // Change current node
        curr = curr->next;
    }
}

void printList(struct Node* head) {
    struct Node* curr = head;
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
    printf("\n");
}

struct Node* createNode(int new_data) {
    struct Node* new_node = 
    	(struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = new_node->child = NULL;
    return new_node;
}

int main() {
  
  	//Linked List - 
  	// 1 -> 2 -> 3
	// |    |   
	// 4 -> 5   6
	// |
	// 7
  
    struct Node *head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->child = createNode(4);
    head->child->next = createNode(5);
    head->next->next->child = createNode(6);
    head->child->child = createNode(7);

    flattenList(head);
    printList(head);

    return 0;
}
Java 
// Java Program to flatten list with 
// next and child pointers

class Node {
    int data;
    Node next, child;

    Node(int x) {
        data = x;
        next = null;
      	child = null;
    }
}

public class GfG {

    // function that flattens 
    // a multilevel linked list
    static void flattenList(Node head) {

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

        // Find tail node of first level
        Node tail = head;
        while (tail.next != null)
            tail = tail.next;

        // One by one traverse through all nodes of first level 
        Node curr = head;

        while (curr != null) {

            // If current node has a child
            if (curr.child != null) {

                // then append the child at the end of current list
                tail.next = curr.child;

                // and update the tail to new last node
                Node tmp = curr.child;
                while (tmp.next != null)
                    tmp = tmp.next;
                tail = tmp;

                // Remove link between curr and child node
                curr.child = null;
            }

            // Change current node
            curr = curr.next;
        }
    }

    static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.data + " ");
            curr = curr.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
      
      	//Linked List - 
      	// 1 -> 2 -> 3
		// |    |   
		// 4 -> 5   6
		// |
		// 7
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.child = new Node(4);
        head.child.next = new Node(5);
        head.next.next.child = new Node(6);
        head.child.child = new Node(7);

        flattenList(head);
        printList(head);
    }
}
Python 
# Python Program to flatten list with 
# next and child pointers

class Node:
    def __init__(self, new_value):
        self.data = new_value
        self.next = None
        self.child = None

# function that flattens 
# a multilevel linked list
def flatten_list(head):

    # Base case
    if head is None:
        return

    # Find tail node of first level
    tail = head
    while tail.next is not None:
        tail = tail.next

    # traverse through all nodes of first level 
    curr = head

    while curr != None:

        # If current node has a child
        if curr.child is not None:

            # then append the child at the end of current list
            tail.next = curr.child

            # and update the tail to new last node
            tmp = curr.child
            while tmp.next is not None:
                tmp = tmp.next
            tail = tmp

            # Remove link between curr and child node
            curr.child = None

        # Change current node
        curr = curr.next

def print_list(head):
    curr = head
    while curr is not None:
        print(curr.data, end=" ")
        curr = curr.next
    print()

if __name__ == "__main__":
  
  	#Linked List - 
  	# 1 -> 2 -> 3
	# |    |   
	# 4 -> 5   6
	# |
	# 7
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.child = Node(4)
    head.child.next = Node(5)
    head.next.next.child = Node(6)
    head.child.child = Node(7)

    flatten_list(head)
    print_list(head)
C# 
// C# Program to flatten list with 
// next and child pointers
using System;

class Node {
    public int data;
    public Node next, child;

    public Node(int x) {
        data = x;
        next = null;
      	child = null;
    }
}

class GfG {

    // function that flattens 
    // a multilevel linked list
    static void FlattenList(Node head) {

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

        // Find tail node of first level
        Node tail = head;
        while (tail.next != null)
            tail = tail.next;

        // traverse through all nodes of first level 
        Node curr = head;

        while (curr != null) {

            // If current node has a child
            if (curr.child != null) {

                // then append the child at the end of current list
                tail.next = curr.child;

                // and update the tail to new last node
                Node tmp = curr.child;
                while (tmp.next != null)
                    tmp = tmp.next;
                tail = tmp;

                // Remove link between curr and child node
                curr.child = null;
            }

            // Change current node
            curr = curr.next;
        }
    }

    static void PrintList(Node head) {
        Node curr = head;
        while (curr != null) {
            Console.Write(curr.data + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main(string[] args) {
      
      	//Linked List - 
      	// 1 -> 2 -> 3
		// |    |   
		// 4 -> 5   6
		// |
		// 7
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.child = new Node(4);
        head.child.next = new Node(5);
        head.next.next.child = new Node(6);
        head.child.child = new Node(7);

        FlattenList(head);
        PrintList(head);
    }
}
JavaScript 
// JavaScript Program to flatten list with 
// next and child pointers

class Node {
    constructor(new_value) {
        this.data = new_value;
        this.next = null;
        this.child = null;
    }
}

// function that flattens 
// a multilevel linked list
function flattenList(head) {

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

    // Find tail node of first level
    let tail = head;
    while (tail.next !== null)
        tail = tail.next;

    // traverse through all nodes of first level 
    let curr = head;

    while (curr !== null) {

        // If current node has a child
        if (curr.child !== null) {

            // then append the child at the end of current list
            tail.next = curr.child;

            // and update the tail to new last node
            let tmp = curr.child;
            while (tmp.next !== null)
                tmp = tmp.next;
            tail = tmp;

            // Remove link between curr and child node
            curr.child = null;
        }

        // Change current node
        curr = curr.next;
    }
}

function printList(head) {
    let curr = head;
    while (curr !== null) {
        console.log(curr.data);
        curr = curr.next;
    }
}

//Linked List - 
// 1 -> 2 -> 3
// |    |   
// 4 -> 5   6
// |
// 7
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.child = new Node(4);
head.child.next = new Node(5);
head.next.next.child = new Node(6);
head.child.child = new Node(7);

flattenList(head);
printList(head);

Output
1 2 3 4 5 6 7

Time Complexity: O(n), where n is the number of nodes in the list.
Auxiliary Space: O(1)


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