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copied!<p>Here is a direct translation of the <code>bftrav</code> Haskell function from the Wikipedia article. Note that it uses a <code>letrec</code> macro I've just written -- see <a href="http://gist.github.com/486880" rel="nofollow noreferrer">this Gist</a> for the latest version.</p> <p>I've changed your definition of <code>my-tree</code> to read thus:</p> <pre><code>(def my-tree (struct tree 1 (struct tree 2) (struct tree 3 (struct tree 4) (struct tree 5)))) </code></pre> <p>Also, my <code>leaf?</code> function assumes that we're only dealing with proper two-way branches and leaf nodes (so a <code>nil</code> on the <code>:left</code> branch implies a <code>nil</code> on the <code>:right</code> branch); it shouldn't be two difficult to change this to handle single-child "branches":</p> <pre><code>(defn leaf? [t] (nil? (:left t))) </code></pre> <p>The code for <code>bftrav</code> is as follows:</p> <pre><code>(defn bftrav [t] (letrec [queue (lazy-seq (cons t (trav 1 queue))) trav (fn [l q] (lazy-seq (cond (zero? l) nil (leaf? (first q)) (trav (dec l) (rest q)) :else (let [{:keys [left right]} (first q)] (cons left (cons right (trav (inc l) (rest q))))))))] queue)) </code></pre> <p>An example from the REPL:</p> <pre><code>user> (bftrav my-tree) ({:val 1, :left {:val 2, :left nil, :right nil}, :right {:val 3, :left {:val 4, :left nil, :right nil}, :right {:val 5, :left nil, :right nil}}} {:val 2, :left nil, :right nil} {:val 3, :left {:val 4, :left nil, :right nil}, :right {:val 5, :left nil, :right nil}} {:val 4, :left nil, :right nil} {:val 5, :left nil, :right nil}) user> (count (bftrav my-tree)) 5 </code></pre>

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