Convert Doubly Linked List to Array II
<p>You are given an <strong>arbitrary</strong> <code>node</code> from a <strong>doubly linked list</strong>, which contains nodes that have a next pointer and a previous pointer.</p>

<p>Return an integer array which contains the elements of the linked list <strong>in order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = [1,2,3,4,5], node = 5</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2,3,4,5]</span></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = [4,5,6,7,8], node = 8</span></p>

<p><strong>Output:</strong> <span class="example-io">[4,5,6,7,8]</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the given list is in the range <code>[1, 500]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 1000</code></li>
	<li>All nodes have unique <code>Node.val</code>.</li>
</ul>
Array; Linked List; Doubly-Linked List
/**
 * Definition for doubly-linked list.
 * class Node {
 *     int val;
 *     Node* prev;
 *     Node* next;
 *     Node() : val(0), next(nullptr), prev(nullptr) {}
 *     Node(int x) : val(x), next(nullptr), prev(nullptr) {}
 *     Node(int x, Node *prev, Node *next) : val(x), next(next), prev(prev) {}
 * };
 */
class Solution {
public:
    vector<int> toArray(Node* node) {
        while (node && node->prev) {
            node = node->prev;
        }
        vector<int> ans;
        for (; node; node = node->next) {
            ans.push_back(node->val);
        }
        return ans;
    }
};
