注意这是双向链表，要把A把B连接到一起，需要同时改两个指针A->next = B, B->prev = A。

时间复杂度：O(n)
空间复杂度：O(n) / stack

class Solution {
public:
  Node* flatten(Node* head) {
    return solve(head).first;
  }
 private:
  // Flaten head, return new head and new tail.
  pair<Node*, Node*> solve(Node* head) {
    if (!head) return {nullptr, nullptr};
    Node dummy;
    Node* cur = &dummy;
    cur->next = head;
    while (cur->next) {
      cur = cur->next;
      if (cur->child) {
        auto [h, t] = solve(cur->child);
        t->next = cur->next;
        h->prev = cur;
        if (cur->next) {
          cur->next->prev = t;
        }
        cur->next = h;
        cur->child = nullptr;
      }
    }
    return {dummy.next, cur};
  }
};