关于java中关闭流疑问

下面这段代码的作用是压缩,会用到ZipOutputStream

  ZipOutputStream zipOutputStream = null;
    try {
      zipOutputStream = new ZipOutputStream(new FileOutputStream(zipPath));
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

错误放大版1:

 ZipOutputStream zipOutputStream = null;
    try {
        FileOutputStream fileOutputStream =   new FileOutputStream(zipPath);
        throw new RuntimeException();
        zipOutputStream = new ZipOutputStream(fileOutputStream);
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

错误放大版2:

    ZipOutputStream zipOutputStream = null;
    try {
      zipOutputStream = new ZipOutputStream(new FileOutputStream(zipPath),null);
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

这里我有一个疑问,在new FileOutputStream(zipPath)的过程中如果已经打开了这个文件,但是在new ZipOutputStream过程中出错了,这个时候zipOutputStream并没有创建成功,所以在finally中不会调用close方法,造成的结果就是这个文件不会被关闭。
所以这种关闭方式是不是会有漏洞,还是我的理由有问题?