# Daily Coding Challenge #59

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
LeetCode - Flatten a Multilevel Doubly Linked List

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked 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 the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Example 1:

Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation:

The multilevel linked list in the input is as follows:

After flattening the multilevel linked list it becomes:

Example 2:

Output: [1,3,2]
Explanation:

The input multilevel linked list is as follows:

1---2---NULL
|
3---NULL
Example 3:

Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above:

1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL
The serialization of each level is as follows:

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]
To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:

[1,2,3,4,5,6,null]
[null,null,7,8,9,10,null]
[null,11,12,null]
Merging the serialization of each level and removing trailing nulls we obtain:

[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

Constraints:

Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5
*/

/*
// Definition for a Node.
class Node {
public:
int val;
Node* prev;
Node* next;
Node* child;
};
*/

class Solution {
public:
// only do the following logic for child
if(h->child){
// store the next node
Node* next=h->next;
// set the next node as child
h->next=h->child;
// since h->next is updated, we need to update the h->next->prev
h->next->prev=h;
// the original node h does not have child now
h->child=NULL;
// n stores the child
Node* n=h->next;
// traverse each child node till the end
while(n->next) n=n->next;
// the last node of the child now points to the next of the original node h
n->next=next;
// if next node is not NULL, we need to update the prev
if(next) next->prev=n;
}
}
}
};

// Step 1:
// h
// 1 - 2 - 3 - 4 - 5 - null
//     |
//     6 - 7 - 8 - null
//             |
//             9 - 10 - null

// Step 2:
//         h
// 1 - 2 - 6 - 7 - 8 - 3 - 4 - 5 - null
//                 |
//                 9 - 10 - null

// Step 3:
//                     h
// 1 - 2 - 6 - 7 - 8 - 9 - 10 - 3 - 4 - 5 - null


The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

Posted on by:

### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.