430 Flatten a Multilevel Doubly Linked List 扁平化多级双向链表
Description:
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:
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]
Explanation:
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]
Explanation:
The input multilevel linked list is as follows:
1---2---NULL
|
3---NULL
Example 3:
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:
[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Constraints:
The number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5
题目描述:
多级双向链表中，除了指向下一个节点和前一个节点指针之外，它还有一个子链表指针，可能指向单独的双向链表。这些子列表也可能会有一个或多个自己的子项，依此类推，生成多级数据结构，如下面的示例所示。
给你位于列表第一级的头节点，请你扁平化列表，使所有结点出现在单级双链表中。
示例 :
示例 1：
输入：head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
输出：[1,2,3,7,8,11,12,9,10,4,5,6]
解释：
输入的多级列表如下图所示：
扁平化后的链表如下图：
示例 2：
输入：head = [1,2,null,3]
输出：[1,3,2]
解释：
输入的多级列表如下图所示：
1---2---NULL
|
3---NULL
示例 3：
输入：head = []
输出：[]
如何表示测试用例中的多级链表？
以 示例 1 为例：
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL
序列化其中的每一级之后：
[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]
为了将每一级都序列化到一起，我们需要每一级中添加值为 null 的元素，以表示没有节点连接到上一级的上级节点。
[1,2,3,4,5,6,null]
[null,null,7,8,9,10,null]
[null,11,12,null]
合并所有序列化结果，并去除末尾的 null 。
[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
提示：
节点数目不超过 1000
1 <= Node.val <= 10^5
思路:
按照题目要求, 每次到有孩子节点时就要保存当前节点的下一节点
所以可以考虑用栈来存储节点
注意有孩子节点的需要将孩子节点置空
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/*
// Definition for a Node.
class Node {
public:
int val;
Node* prev;
Node* next;
Node* child;
};
*/
class Solution
{
public:
Node* flatten(Node* head)
{
Node* cur = head;
stack<Node*> s;
while (cur)
{
if (cur -> child)
{
if (cur -> next) s.push(cur -> next);
cur -> next = cur -> child;
cur -> child -> prev = cur;
cur -> child = nullptr;
}
else if (!cur -> next and !s.empty())
{
cur -> next = s.top();
s.pop();
cur -> next -> prev = cur;
}
cur = cur -> next;
}
return head;
}
};
Java:
/*
// Definition for a Node.
class Node {
public int val;
public Node prev;
public Node next;
public Node child;
};
*/
class Solution {
public Node flatten(Node head) {
Node cur = head;
Stack<Node> stack = new Stack<>();
while (cur != null) {
if (cur.child != null) {
if (cur.next != null) stack.push(cur.next);
cur.child.prev = cur;
cur.next = cur.child;
cur.child = null;
} else if (cur.next == null && !stack.isEmpty()) {
cur.next = stack.pop();
cur.next.prev = cur;
}
cur = cur.next;
}
return head;
}
}
Python:
"""
# 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':
cur, stack = head, []
while cur:
if cur.child:
if cur.next:
stack.append(cur.next)
cur.child.prev, cur.next, cur.child = cur, cur.child, None
elif not cur.next and stack:
cur.next, cur.next.prev = stack.pop(), cur
cur = cur.next
return head
