links method

List<Link<T>> links()

返回当前节点的后续的所有Link

Implementation

List<Link<T>> links() {
  List<Link<T>> links = [];
  each((node, index, startNode) {
    if (node != this && node.parent != null) {
      links.add(Link(node.parent!, node));
    }
    return false;
  });
  return links;
}