leetcode 430 扁平化多级双向链表
z
 2019-07-01 14:51:33  

您将获得一个双向链表，除了下一个和前一个指针之外，它还有一个子指针，可能指向单独的双向链表。这些子列表可能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。扁平化列表，使所有结点出现在单级双链表中。您将获得列表第一级的头部。

示例:

输入:
1—2—3—4—5—6–NULL
|
7—8—9—10–NULL
|
11–12–NULL

输出:
1-2-3-7-8-11-12-9-10-4-5-6-NULL

multilevellinkedlist

multilevellinkedlistflattened

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"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""
class Solution(object):
    def flatten(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if not head:
            return head
        pres = []
        dump = Node(-1)
        pre = dump
        cur = None
        pres.append(head)
        while pres:
            if not cur:
                cur = pres.pop(-1)
            while cur and not cur.child:
                pre.next = cur
                cur.prev = pre
                pre = pre.next
                cur = cur.next
            if cur:
                pre.next = cur
                cur.prev = pre
                pre = pre.next
                if cur.next:
                    pres.append(cur.next)
                pres.append(cur.child)
                cur.child = None
                cur = None
        dump.next.prev = None
        return dump.next
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"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""
class Solution(object):
    def flatten(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if not head:
            return head
        stack = []
        prev = None
        stack.append(head)
        while stack:
            cur = stack.pop(-1)
            if cur.next:
                stack.append(cur.next)
            if cur.child:
                stack.append(cur.child)
                cur.child = None
            if prev:
                prev.next = cur
                cur.prev = prev
                prev = prev.next
            else:
                prev = cur
        return head