[spiralx]

You can do it either way in XPath thanks to how you can use a path expression and/or predicates almost everywhere in a query

  # Find all elements li and select the parent element for each
  //li/.. 

  # Find all element nodes with a child element named li
  //*[li]

  # Non-abbreviated queries
  /descendant::li/parent::*
  /descendant::*[child::li]

  # CSS using :has
  :has(> li)