111 lines
2.4 KiB
Markdown
111 lines
2.4 KiB
Markdown
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---
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layout: page
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title: Flatten a Multilevel Doubly Linked List
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last_solved: 2020-07-10
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category: linked list
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leetcode_url: https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/
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status: Attempted
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---
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Problem
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-------
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```
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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.
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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.
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Example 1:
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Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
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Output: [1,2,3,7,8,11,12,9,10,4,5,6]
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Explanation:
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The multilevel linked list in the input is as follows:
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After flattening the multilevel linked list it becomes:
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Example 2:
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Input: head = [1,2,null,3]
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Output: [1,3,2]
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Explanation:
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The input multilevel linked list is as follows:
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1---2---NULL
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3---NULL
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Example 3:
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Input: head = []
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Output: []
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How multilevel linked list is represented in test case:
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We use the multilevel linked list from Example 1 above:
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1---2---3---4---5---6--NULL
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7---8---9---10--NULL
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11--12--NULL
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The serialization of each level is as follows:
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[1,2,3,4,5,6,null]
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[7,8,9,10,null]
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[11,12,null]
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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:
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[1,2,3,4,5,6,null]
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[null,null,7,8,9,10,null]
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[null,11,12,null]
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Merging the serialization of each level and removing trailing nulls we obtain:
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[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
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```
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Solution
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----------
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{% highlight python %}
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def flatten(self, head: 'Node') -> 'Node':
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if not head: return head
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dummy = Node(0)
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curr = dummy
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stack = [head]
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while stack:
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tmp = stack.pop()
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if tmp.next: stack.append( tmp.next )
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if tmp.child: stack.append( tmp.child )
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print( tmp.val )
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curr.next = tmp
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tmp.prev = curr
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tmp.child = None # no child pointers
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curr = tmp # move ahead
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dummy.next.prev = None
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return dummy.next
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{% endhighlight %}
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![image1]()
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