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    copied!<p>As others have mentioned, build a hashtable/map of objects to a list of their (direct) children.</p> <p>From there you can easily lookup a list of direct children of your "target object", and then for each object in the list, repeat the process.</p> <p>Here's how I did it in Java and using generics, with a queue instead of any recursion:</p> <pre><code>public static Set&lt;Node&gt; findDescendants(List&lt;Node&gt; allNodes, Node thisNode) { // keep a map of Nodes to a List of that Node's direct children Map&lt;Node, List&lt;Node&gt;&gt; map = new HashMap&lt;Node, List&lt;Node&gt;&gt;(); // populate the map - this is O(n) since we examine each and every node // in the list for (Node n : allNodes) { Node parent = n.getParent(); if (parent != null) { List&lt;Node&gt; children = map.get(parent); if (children == null) { // instantiate list children = new ArrayList&lt;Node&gt;(); map.put(parent, children); } children.add(n); } } // now, create a collection of thisNode's children (of all levels) Set&lt;Node&gt; allChildren = new HashSet&lt;Node&gt;(); // keep a "queue" of nodes to look at List&lt;Node&gt; nodesToExamine = new ArrayList&lt;Node&gt;(); nodesToExamine.add(thisNode); while (nodesToExamine.isEmpty() == false) { // pop a node off the queue Node node = nodesToExamine.remove(0); List&lt;Node&gt; children = map.get(node); if (children != null) { for (Node c : children) { allChildren.add(c); nodesToExamine.add(c); } } } return allChildren; } </code></pre> <p>The expected execution time is something between O(n) and O(2n), if I remember how to calculate that right. You're guaranteed to look at every node in the list, plus a few more operations to find all of the descendants of your node - in the worst case (if you run the algorithm on the root node) you are looking at every node in the list twice.</p>
 

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