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.
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] Explanation:
The multilevel linked list in the input is as follows:
After flattening the multilevel linked list it becomes:
Input: head = [1,2,null,3] Output: [1,3,2] Explanation: The input multilevel linked list is as follows: 1---2---NULL | 3---NULL
Input: head =  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:
- Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5
1. Coding Skill
主要考验的是编程能力，看能不能够 bug free 了！
# Time: O(n) # Space: O(1) """ # Definition for a Node. 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': if not head: return head cur = head while cur: # CASE 1: if no child, proceed if not cur.child: cur = cur.next continue # CASE 2: got child, find the tail of the child and link it to cur.next tmp = cur.child # find the tail of the child while tmp.next: tmp = tmp.next # connect tail with cur.next, if it is not None tmp.next = cur.next if cur.next: cur.next.prev = tmp cur.next = cur.child cur.child.prev = cur cur.child = None return head # 22/22 cases passed (36 ms) # Your runtime beats 72.4 % of python3 submissions # Your memory usage beats 100 % of python3 submissions (12.8 MB)
2. Stack + two pointers
总有一个先处理 child 后处理 next 的操作，就可以用 stack 来存。
class Solution: def flatten(self, head: 'Node') -> 'Node': if not head: return head stack = [head] prev = Node(0) while stack: node = stack.pop() node.prev = prev prev.next = node prev = node if node.next: stack.append(node.next) if node.child: stack.append(node.child) node.child = None head.prev = None return head # 22/22 cases passed (32 ms) # Your runtime beats 87.74 % of python3 submissions # Your memory usage beats 100 % of python3 submissions (12.8 MB)