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    <p>The answer is : <strong>you can recover a binary search tree from its pre order traversal</strong>. </p> <p>I'm not sure what is your mathematical background so please ask if you need more explanation. </p> <p>For simplicity, I'm assuming that the node are labeled by the integer 1,2... n where n is the number of node. Then the pre-order traversal of the tree t gives gives you a permutation of <code>[n] = {1,2,...,n}</code> which have a particular property: each time you have a letter b in your permutation, you can't find two consecutive letter <code>ca</code> after the <code>b</code> in the permutation such that <code>a&lt;b&lt;c</code>. Such a permutation is said to <em>avoid</em> the pattern <code>b-ac</code> (the - stand for an arbitrary number of letters). </p> <p>For example, 4 2 1 3 avoids b-ac whereas 3 1 4 2 doesn't because 3 - 4 2.</p> <p>This is actually an equivalence: A permutation is the pre order reading of a tree iff is avoid b-ac. </p> <p>It is know that there are as many trees of size n as permutation avoiding b-ac so this is a bijection. Their number are know as Catalan number. You probably can find this as an exercice of Stanley's book "enumerative combinatorics".</p> <p><strong>Here is a more algorithmic explanation</strong>:</p> <pre><code>RecTree: Recovering a tree from is Pre-order traversal: input: list l output: tree t b &lt;- l[0] find an index i such that - for 1&lt;=j&lt;=i then l[j] &lt; b and - for i&lt;j&lt;=n then l[j] &gt; b if there isn't exists such an index return Failure else return Node(key=b, RecTree(l[1..i]), RecTree(l[i+1..n])) </code></pre> <p>As a consequence </p> <p><strong>Two binary search trees are equal if and only if they have the same pre order traversal</strong> </p> <p>Does it makes sense to you ?</p> <p>Some more references </p> <ul> <li><a href="http://oeis.org/A000108," rel="nofollow">Catalan number in the On-Line Encyclopedia of Integer Sequences</a></li> <li><a href="http://combinatorics.cis.strath.ac.uk/download/Cl01__Generalized_Pattern.pdf" rel="nofollow">Anders Claesson Generalized Pattern Avoidance, European Journal of Combinatorics 22 (2001) 961–971</a></li> </ul>
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    1. COThanks for the great answer. Could you please elaborate the part, you can't find two consecutive letter ca after the b in the permutation such that aavoid the pattern b-ac (the - stand for an arbitrary number of letters). May be I couldn't get what do you mean by a, b,c. Like a < b < c ??
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    2. COIt is clearer like that ? The markup garbled my text.
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    3. COGreat, I got the point, thanks a ton. :)
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