leetcode-430 扁平化多级双向链表

解题思路：需要将层级化结果转为扁平化；当前节点head，如果有child的化，需要遍历child，遍历child的过程就是一个递归的过程，递归时只需要将节点最后一个链表while(child->next) child = child->next和head->next进行相连即可（迭代方法：需要一个挨着一个进行遍历）；如果没有child的化，只需要head = head->next即可；

递归

Node* flatten(Node* head) {
    Node* res = new Node(0);
    res->next = head;
    while (head != NULL) {
        if (head->child == NULL) {
            head = head->next;
        } else {
            Node* temp = head->next;
            Node* child_head = flatten(head->child);

            //子节点和head进行相连接
            head->next = child_head;
            child_head->prev = head;
            head->child = NULL;
            while (head->next != NULL) {
                head = head->next;
            }
            if (temp != NULL) {
                head->next = temp;
            }
            if (temp != NULL) {
                temp->prev = head;
            }
            head = temp;
        }
    }
    return res->next;
}

迭代

Node* flatten(Node* head) {
    Node* res = new Node(-1);
    res->next = head;
    while (head != NULL) {
        if (head->child == NULL) {
            head = head->next;
        } else {
            Node* temp = head->next;
            Node* child_head = head->child;
            head->child = NULL;
            head->next = child_head;
            child_head->prev = head;
            while (child_head->next != NULL) {
                child_head = child_head->next;
            }
            child_head->next = temp;
            if (temp != NULL) {
                temp->prev = child_head;
            }
            head = head->next;
        }
    }
    return res->next;
}