如何在此函数中获取路径

如何在此函数中获取路径？这个函数只给我名字，但我想获取路径该怎么做？

   function getDirectory($path = '.', $level = 1) {
       $result = array();
       $ignore = array('nbproject', 'src', '.', '..');
       $dh = @opendir($path);
       $i = 0;
       while ($file = readdir($dh)) {
           if (!in_array($file, $ignore)) {
             if (is_dir($path . '/' . $file)) {
                 $level++;
                 $singleResult = array('title' => $file, 'isFolder' => true, 'children' => getDirectory($path . '/' . $file, $level), 'key' => '/' . $file);
                 $result[] = $singleResult;
               }
            }
            $i++;
          }
          closedir($dh);
           return $result;
 }
 $dir = "../UserUpload/Documents/source";
 $kevin = getDirectory($dir);

这个函数给了我这样的数组。我也想在这个数组中获取路径该怎么做？

array (size=3)
  0 => 
    array (size=4)
      'title' => string 'mst146' (length=6)
      'isFolder' => boolean true
      'children' => 
      'key' => string '/mst146' (length=7)
  1 => 
    array (size=4)
      'title' => string 't124' (length=4)
      'isFolder' => boolean true
      'children' => 
      'key' => string '/t124' (length=5)
  2 => 
    array (size=4)
      'title' => string 'test' (length=4)
      'isFolder' => boolean true
      'children' => 
      'key' => string '/test' (length=5)