LeetCode 430. 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:

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]

The multilevel linked list in the input is as follows:

After flattening the multilevel linked list it becomes:

Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
The input multilevel linked list is as follows:


Example 3:

Input: head = []
Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above:

The serialization of each level is as follows:

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:

Merging the serialization of each level and removing trailing nulls we obtain:



  • 使用循环,从头开始遍历,将子链看成一个整体,将整体向上移动一层。
  • 继续向后遍历,如果遇到子链,将其看作一层,将其整体向上移动,如此循环下去,直至结束。
# -*- coding: utf-8 -*-
# @Author:             何睿
# @Create Date:        2019-12-29 10:57:03
# @Last Modified by:   何睿
# @Last Modified time: 2019-12-29 11:05:17

class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child

class Solution:
    def flatten(self, head: 'Node') -> 'Node':
        current = head
        while current:
            if current.child:
                _next = current.next  # 当前节点的后一个节点
                last = current.child  # 当前节点的自节点
                while last.next:
                    last = last.next  # 如果有子节点,找到子节点的最后一个节点
                current.next = current.child
                current.next.prev = current  # 将自节点向上提高一层
                current.child = None
                last.next = _next
                if _next:
                    _next.prev = last
            current = current.next

        return head

源代码文件在 这里 。

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