题目：

题解：

class Solution {public Node flatten(Node head) {dfs(head);return head;}public Node dfs(Node node) {Node cur = node;// 记录链表的最后一个节点Node last = null;while (cur != null) {Node next = cur.next;//  如果有子节点，那么首先处理子节点if (cur.child != null) {Node childLast = 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 != null) {childLast.next = next;next.prev = childLast;}// 将 child 置为空cur.child = null;last = childLast;} else {last = cur;}cur = next;}return last;}
}