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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。

给定位于列表第一级的头节点,请扁平化列表,即将这样的多级双向链表展平成普通的双向链表,使所有结点出现在单级双链表中。

 

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?

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

 

注意:本题与主站 430 题相同: https://leetcode.cn/problems/flatten-a-multilevel-doubly-linked-list/

解法

仔细观察一下这个结构,不难发现其实就是前序遍历二叉树

Python3

"""
# 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 head is None:
            return None
        dummy = Node()
        tail = dummy

        def preOrder(node: 'Node'):
            nonlocal tail
            if node is None:
                return
            next = node.next
            child = node.child
            tail.next = node
            node.prev = tail
            tail = tail.next
            node.child = None
            preOrder(child)
            preOrder(next)

        preOrder(head)
        dummy.next.prev = None
        return dummy.next

Java

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

class Solution {
    private Node dummy = new Node();
    private Node tail = dummy;

    public Node flatten(Node head) {
        if (head == null) {
            return null;
        }
        preOrder(head);
        dummy.next.prev = null;
        return dummy.next;
    }

    private void preOrder(Node node) {
        if (node == null) {
            return;
        }
        Node next = node.next;
        Node child = node.child;
        tail.next = node;
        node.prev = tail;
        tail = tail.next;
        node.child = null;
        preOrder(child);
        preOrder(next);
    }
}

C++

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

class Solution {
public:
    Node* flatten(Node* head) {
        flattenGetTail(head);
        return head;
    }

    Node* flattenGetTail(Node* head) {
        Node* cur = head;
        Node* tail = nullptr;

        while (cur) {
            Node* next = cur->next;
            if (cur->child) {
                Node* child = cur->child;
                Node* childTail = flattenGetTail(cur->child);

                cur->child = nullptr;
                cur->next = child;
                child->prev = cur;
                childTail->next = next;

                if (next)
                    next->prev = childTail;

                tail = childTail;
            } else {
                tail = cur;
            }

            cur = next;
        }

        return tail;
    }
};

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