题目：

题解：

class Solution:
    def flatten(self, head: "Node") -> "Node":
        def dfs(node: "Node") -> "Node":
            cur = node
            # 记录链表的最后一个节点
            last = None

            while cur:
                nxt = cur.next
                # 如果有子节点，那么首先处理子节点
                if cur.child:
                    child_last = dfs(cur.child)
                    
                    nxt = cur.next
                    # 将 node 与 child 相连
                    cur.next = cur.child
                    cur.child.prev = cur

                    # 如果 nxt 不为空，就将 last 与 nxt 相连
                    if nxt:
                        child_last.next = nxt
                        nxt.prev = child_last

                    # 将 child 置为空
                    cur.child = None
                    last = child_last
                else:
                    last = cur
                cur = nxt

            return last

        dfs(head)
        return head