Flatten a multilevel linked list

20 Jun, 2022
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, and so on, to produce a multilevel data structure, as shown in below figure. You are given the head of the first level of the list. Flatten the list so that all the nodes appear in a single-level linked list. You need to flatten the list in way that all nodes at the first level should come first, then nodes of second level, and so on.Each node is a C struct with the following definition.
C
Java
Python3
C#
Javascript
struct List{    int data;    struct List *next;    struct List *child;};
static class List{    public int data;    public List next;    public List child;}; // This code is contributed by pratham76
# A linked list node has data,# next pointer and child pointerclass Node:    def __init__(self, data):        self.data = data        self.next = None        self.child = None                 # This code contributed by umadevi9616
static class List{    public int data;    public List next;    public List child;}; // This code is contributed by rutvik_56
<script>class List{     constructor(){         this.data = 0;         this.next = null;         this.child = null;     } }// This code contributed by aashish1995</script>
The above list should be converted to 10->5->12->7->11->4->20->13->17->6->2->16->9->8->3->19->15 
The problem clearly say that we need to flatten level by level. The idea of solution is, we start from first level, process all nodes one by one, if a node has a child, then we append the child at the end of list, otherwise we don’t do anything. After the first level is processed, all next level nodes will be appended after first level. Same process is followed for the appended nodes. 
1) Take "cur" pointer, which will point to head of the first level of the list
2) Take "tail" pointer, which will point to end of the first level of the list
3) Repeat the below procedure while "curr" is not NULL.
    I) if current node has a child then
    a) append this new child list to the "tail"
        tail->next = cur->child
    b) find the last node of new child list and update "tail"
        tmp = cur->child;
        while (tmp->next != NULL)
            tmp = tmp->next;
        tail = tmp;
    II) move to the next node. i.e. cur = cur->next
Following is the implementation of the above algorithm. 
C++
C
Java
Python3
C#
// C++ Program to flatten list with// next and child pointers#include <bits/stdc++.h>using namespace std; // Macro to find number of elements in array#define SIZE(arr) (sizeof(arr)/sizeof(arr[0])) // A linked list node has data,// next pointer and child pointerclass Node{    public:    int data;    Node *next;    Node *child;}; // A utility function to create a linked list// with n nodes. The data of nodes is taken// from arr[]. All child pointers are set as NULLNode *createList(int *arr, int n){    Node *head = NULL;    Node *p;     int i;    for (i = 0; i < n; ++i)    {        if (head == NULL)            head = p = new Node();        else        {            p->next = new Node();            p = p->next;        }        p->data = arr[i];        p->next = p->child = NULL;    }    return head;} // A utility function to print// all nodes of a linked listvoid printList(Node *head){    while (head != NULL)    {        cout << head->data << " ";        head = head->next;    }    cout<<endl;} // This function creates the input// list. The created list is same// as shown in the above figureNode *createList(void){    int arr1[] = {10, 5, 12, 7, 11};    int arr2[] = {4, 20, 13};    int arr3[] = {17, 6};    int arr4[] = {9, 8};    int arr5[] = {19, 15};    int arr6[] = {2};    int arr7[] = {16};    int arr8[] = {3};     /* create 8 linked lists */    Node *head1 = createList(arr1, SIZE(arr1));    Node *head2 = createList(arr2, SIZE(arr2));    Node *head3 = createList(arr3, SIZE(arr3));    Node *head4 = createList(arr4, SIZE(arr4));    Node *head5 = createList(arr5, SIZE(arr5));    Node *head6 = createList(arr6, SIZE(arr6));    Node *head7 = createList(arr7, SIZE(arr7));    Node *head8 = createList(arr8, SIZE(arr8));      /* modify child pointers to    create the list shown above */    head1->child = head2;    head1->next->next->next->child = head3;    head3->child = head4;    head4->child = head5;    head2->next->child = head6;    head2->next->next->child = head7;    head7->child = head8;      /* Return head pointer of first    linked list. Note that all nodes are    reachable from head1 */    return head1;} /* The main function that flattensa multilevel linked list */void flattenList(Node *head){    /*Base case*/    if (head == NULL)    return;     Node *tmp;     /* Find tail node of first level linked list */    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    Node *cur = head;    while (cur != tail)    {        // If current node has a child        if (cur->child)        {            // then append the child at the end of current list            tail->next = cur->child;             // and update the tail to new last node            tmp = cur->child;            while (tmp->next)                tmp = tmp->next;            tail = tmp;        }         // Change current node        cur = cur->next;    }} // Driver codeint main(void){    Node *head = NULL;    head = createList();    flattenList(head);    printList(head);    return 0;} // This code is contributed by rathbhupendra
// Program to flatten list with next and child pointers#include <stdio.h>#include <stdlib.h> // Macro to find number of elements in array#define SIZE(arr) (sizeof(arr)/sizeof(arr[0])) // A linked list node has data, next pointer and child pointerstruct Node{    int data;    struct Node *next;    struct Node *child;}; // A utility function to create a linked list with n nodes. The data// of nodes is taken from arr[].  All child pointers are set as NULLstruct Node *createList(int *arr, int n){    struct Node *head = NULL;    struct Node *p;     int i;    for (i = 0; i < n; ++i) {        if (head == NULL)            head = p = (struct Node *)malloc(sizeof(*p));        else {            p->next = (struct Node *)malloc(sizeof(*p));            p = p->next;        }        p->data = arr[i];        p->next = p->child = NULL;    }    return head;} // A utility function to print all nodes of a linked listvoid printList(struct Node *head){    while (head != NULL) {        printf("%d ", head->data);        head = head->next;    }    printf("\n");} // This function creates the input list.  The created list is same// as shown in the above figurestruct Node *createList(void){    int arr1[] = {10, 5, 12, 7, 11};    int arr2[] = {4, 20, 13};    int arr3[] = {17, 6};    int arr4[] = {9, 8};    int arr5[] = {19, 15};    int arr6[] = {2};    int arr7[] = {16};    int arr8[] = {3};     /* create 8 linked lists */    struct Node *head1 = createList(arr1, SIZE(arr1));    struct Node *head2 = createList(arr2, SIZE(arr2));    struct Node *head3 = createList(arr3, SIZE(arr3));    struct Node *head4 = createList(arr4, SIZE(arr4));    struct Node *head5 = createList(arr5, SIZE(arr5));    struct Node *head6 = createList(arr6, SIZE(arr6));    struct Node *head7 = createList(arr7, SIZE(arr7));    struct Node *head8 = createList(arr8, SIZE(arr8));      /* modify child pointers to create the list shown above */    head1->child = head2;    head1->next->next->next->child = head3;    head3->child = head4;    head4->child = head5;    head2->next->child = head6;    head2->next->next->child = head7;    head7->child = head8;      /* Return head pointer of first linked list.  Note that all nodes are       reachable from head1 */    return head1;} /* The main function that flattens a multilevel linked list */void flattenList(struct Node *head){    /*Base case*/    if (head == NULL)       return;     struct Node *tmp;     /* Find tail node of first level linked list */    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 *cur = head;    while (cur != tail)    {        // If current node has a child        if (cur->child)        {            // then append the child at the end of current list            tail->next = cur->child;             // and update the tail to new last node            tmp = cur->child;            while (tmp->next)                tmp = tmp->next;            tail = tmp;        }         // Change current node        cur = cur->next;    }} // A driver program to test above functionsint main(void){    struct Node *head = NULL;    head = createList();    flattenList(head);    printList(head);    return 0;}
// Java program to flatten linked list with next and child pointers class LinkedList {         static Node head;         class Node {                 int data;        Node next, child;                 Node(int d) {            data = d;            next = child = null;        }    }     // A utility function to create a linked list with n nodes. The data    // of nodes is taken from arr[].  All child pointers are set as NULL    Node createList(int arr[], int n) {        Node node = null;        Node p = null;                 int i;        for (i = 0; i < n; ++i) {            if (node == null) {                node = p = new Node(arr[i]);            } else {                p.next = new Node(arr[i]);                p = p.next;            }            p.next = p.child = null;        }        return node;    }     // A utility function to print all nodes of a linked list    void printList(Node node) {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }        System.out.println("");    }         Node createList() {        int arr1[] = new int[]{10, 5, 12, 7, 11};        int arr2[] = new int[]{4, 20, 13};        int arr3[] = new int[]{17, 6};        int arr4[] = new int[]{9, 8};        int arr5[] = new int[]{19, 15};        int arr6[] = new int[]{2};        int arr7[] = new int[]{16};        int arr8[] = new int[]{3};         /* create 8 linked lists */        Node head1 = createList(arr1, arr1.length);        Node head2 = createList(arr2, arr2.length);        Node head3 = createList(arr3, arr3.length);        Node head4 = createList(arr4, arr4.length);        Node head5 = createList(arr5, arr5.length);        Node head6 = createList(arr6, arr6.length);        Node head7 = createList(arr7, arr7.length);        Node head8 = createList(arr8, arr8.length);         /* modify child pointers to create the list shown above */        head1.child = head2;        head1.next.next.next.child = head3;        head3.child = head4;        head4.child = head5;        head2.next.child = head6;        head2.next.next.child = head7;        head7.child = head8;         /* Return head pointer of first linked list.  Note that all nodes are         reachable from head1 */        return head1;    }     /* The main function that flattens a multilevel linked list */    void flattenList(Node node) {                 /*Base case*/        if (node == null) {            return;        }                 Node tmp = null;         /* Find tail node of first level linked list */        Node tail = node;        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        Node cur = node;        while (cur != tail) {             // If current node has a child            if (cur.child != null) {                 // then append the child at the end of current list                tail.next = cur.child;                 // and update the tail to new last node                tmp = cur.child;                while (tmp.next != null) {                    tmp = tmp.next;                }                tail = tmp;            }             // Change current node            cur = cur.next;        }    }         public static void main(String[] args) {        LinkedList list = new LinkedList();        head = list.createList();        list.flattenList(head);        list.printList(head);    }} // This code has been contributed by Mayank Jaiswal
# Python3 Program to flatten list with# next and child pointers # A linked list node has data,# next pointer and child pointerclass Node:    def __init__(self, data):        self.data = data        self.next = None        self.child = None # Return Nodedef newNode(data):    return Node(data) # The main function that flattens# a multilevel linked listdef flattenlist(head):         # Base case    if not head:        return         # Find tail node of first level linked list    temp = head    while(temp.next != None):        temp = temp.next    currNode = head         # One by one traverse through all nodes    # of first level linked list    # till we reach the tail node    while(currNode != temp):                 # If current node has a child        if(currNode.child):                         # then append the child            # at the end of current list            temp.next = currNode.child                         # and update the tail to new last node            tmp = currNode.child            while(tmp.next):                tmp = tmp.next            temp = tmp                     # Change current node        currNode = currNode.next     # A utility function to print# all nodes of a linked listdef printList(head):    if not head:        return    while(head):        print("{}".format(head.data), end = " ")        head = head.next # Driver codeif __name__=='__main__':         # Child list of 13    child13 = newNode(16)    child13.child = newNode(3)         # Child List of 10    head1 = newNode(4)    head1.next = newNode(20)    head1.next.child = newNode(2) #Child of 20    head1.next.next = newNode(13)    head1.next.next.child = child13         # Child of 9    child9 = newNode(19)    child9.next = newNode(15)     # Child List of 17    child17 = newNode(9)    child17.next = newNode(8)    child17.child = child9     # Child List of 7    head2 = newNode(17)    head2.next = newNode(6)    head2.child = child17         # Main List    head = newNode(10)    head.child = head1    head.next = newNode(5)    head.next.next = newNode(12)    head.next.next.next = newNode(7)    head.next.next.next.child = head2    head.next.next.next.next = newNode(11)     flattenlist(head)     print("\Flattened list is: ", end = "")    printList(head) # This code is contributed by 0_hero
// C# program to flatten linked list// with next and child pointersusing System; public class LinkedList{    static Node head;         class Node    {        public int data;        public Node next, child;                 public Node(int d)        {            data = d;            next = child = null;        }    }     // A utility function to create    // a linked list with n nodes. The data    // of nodes is taken from arr[].    // All child pointers are set as NULL    Node createList(int []arr, int n)    {        Node node = null;        Node p = null;                 int i;        for (i = 0; i < n; ++i)        {            if (node == null)            {                node = p = new Node(arr[i]);            }            else            {                p.next = new Node(arr[i]);                p = p.next;            }            p.next = p.child = null;        }        return node;    }     // A utility function to print    // all nodes of a linked list    void printList(Node node)    {        while (node != null)        {            Console.Write(node.data + " ");            node = node.next;        }        Console.WriteLine("");    }         Node createList()    {        int []arr1 = new int[]{10, 5, 12, 7, 11};        int []arr2 = new int[]{4, 20, 13};        int []arr3 = new int[]{17, 6};        int []arr4 = new int[]{9, 8};        int []arr5 = new int[]{19, 15};        int []arr6 = new int[]{2};        int []arr7 = new int[]{16};        int []arr8 = new int[]{3};         /* create 8 linked lists */        Node head1 = createList(arr1, arr1.Length);        Node head2 = createList(arr2, arr2.Length);        Node head3 = createList(arr3, arr3.Length);        Node head4 = createList(arr4, arr4.Length);        Node head5 = createList(arr5, arr5.Length);        Node head6 = createList(arr6, arr6.Length);        Node head7 = createList(arr7, arr7.Length);        Node head8 = createList(arr8, arr8.Length);         /* modify child pointers to        create the list shown above */        head1.child = head2;        head1.next.next.next.child = head3;        head3.child = head4;        head4.child = head5;        head2.next.child = head6;        head2.next.next.child = head7;        head7.child = head8;         /* Return head pointer of first        linked list. Note that all nodes         are reachable from head1 */        return head1;    }     /* The main function that flattens    a multilevel linked list */    void flattenList(Node node)    {                 /*Base case*/        if (node == null)        {            return;        }                 Node tmp = null;         /* Find tail node of first        level linked list */        Node tail = node;        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        Node cur = node;        while (cur != tail)        {             // If current node has a child            if (cur.child != null)            {                 // then append the child at                // the end of current list                tail.next = cur.child;                 // and update the tail to new last node                tmp = cur.child;                while (tmp.next != null)                {                    tmp = tmp.next;                }                tail = tmp;            }             // Change current node            cur = cur.next;        }    }         // Driver code    public static void Main()    {        LinkedList list = new LinkedList();        head = list.createList();        list.flattenList(head);        list.printList(head);    }} /* This code is contributed PrinciRaj1992 */
10 5 12 7 11 4 20 13 17 6 2 16 9 8 3 19 15 
Time Complexity: Since every node is visited at most twice, the time complexity is O(n) where n is the number of nodes in given linked list.
princiraj1992
rathbhupendra
0_hero
rutvik_56
pratham76
aashish1995
umadevi9616
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