如何在点击链接时关闭全屏菜单？

const menuLinks = document.querySelectorAll('.menu-link');
const fullscreenMenu = document.querySelector('.fullscreen-menu');

menuLinks.forEach(link => {
  link.addEventListener('click', () => {
    // 在这里执行关闭全屏菜单的操作
    fullscreenMenu.classList.remove('active');
  });
});

<div class="fullscreen-menu">
  <ul class="menu">
    <li><a class="menu-link" href="#">链接1</a></li>
    <li><a class="menu-link" href="#">链接2</a></li>
    <li><a class="menu-link" href="#">链接3</a></li>
  </ul>
</div>

.fullscreen-menu {
  position: fixed;
  top: 0;
  left: 0;
  width: 100%;
  height: 100%;
  background-color: #000;
  color: #fff;
  display: none;
}

.fullscreen-menu.active {
  display: block;
}