题目：

题解：

class Solution {
public:Node* flatten(Node* head) {function<Node*(Node*)> dfs = [&](Node* node) {Node* cur = node;// 记录链表的最后一个节点Node* last = nullptr;while (cur) {Node* next = cur->next;//  如果有子节点，那么首先处理子节点if (cur->child) {Node* child_last = dfs(cur->child);next = cur->next;//  将 node 与 child 相连cur->next = cur->child;cur->child->prev = cur;//  如果 next 不为空，就将 last 与 next 相连if (next) {child_last->next = next;next->prev = child_last;}// 将 child 置为空cur->child = nullptr;last = child_last;}else {last = cur;}cur = next;}return last;};dfs(head);return head;}
};