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 a Node.
 * type Node struct {
 *     Val int
 *     Next *Node
 *     Prev *Node
 * }
 */

func toArray(node *Node) (ans []int) {
	for node != nil && node.Prev != nil {
		node = node.Prev
	}
	for ; node != nil; node = node.Next {
		ans = append(ans, node.Val)
	}
	return
}
