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Home/ Questions/Q 494625
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T05:30:49+00:00 2026-05-13T05:30:49+00:00

I am trying to build a directory tree such as how xml trees are

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I am trying to build a directory tree such as how xml trees are represented, in the form of a vector, i can traverse the file system fine using the following snippet but i can’t put my head around how to build a tree structure out of this?


(defn trav [dir]
  (if (.isDirectory dir)
    (do
      (println (.getName dir))
      (doseq [file (.listFiles dir)]
        (if (.isDirectory file)
          (trav file)))      
      )))
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  1. Editorial Team
    Editorial Team
    2026-05-13T05:30:49+00:00Added an answer on May 13, 2026 at 5:30 am

    How about this?

    (defstruct file :file)
    (defstruct dir :file :contents)
    
    (defn file-tree
      [#^File file]
      (if (.isDirectory file)
        (struct dir file (vec (map file-tree (.listFiles file))))
        (struct file file)))

    If you query the resulting map for :file you get the file entry for this node back. If you ask for :contents and get nil, it’s a file. A vector indicates a directory.

    As Carl already said: maybe file-seq is more appropriate.

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