C++ Program For Flattening A Multilevel Linked List - GeeksforGeeks

22 Dec, 2021
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 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 a way that all nodes at the first level should come first, then nodes of the second level, and so on.
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 says that we need to flatten level by level. The idea of a solution is, we start from the first level, process all nodes one by one, if a node has a child, then we append the child at the end of the list, otherwise, we don’t do anything. After the first level is processed, all next-level nodes will be appended after the first level. The same process is followed for the appended nodes. 
1) Take the "cur" pointer, which will point to the head 
        of the first level of the list
2) Take the "tail" pointer, which will point to the end of the 
   first level of the list
3) Repeat the below procedure while "curr" is not NULL.
    I) If the 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 the new child list and update 
       the "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++ 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 pointer class 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 NULL Node *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 list void 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 figure 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     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 flattens   a 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
Output:
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.
Please refer complete article on Flatten a multilevel linked list for more details!
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