Knowing that Semgraph member is a Set should be easier to understand when looking at the collection of roots

Also, add a test of if a node is a root or not

Author: AngledLuffa

src/edu/stanford/nlp/semgraph/SemanticGraph.java

@@ -67,7 +67,7 @@ public class SemanticGraph implements Serializable  {
   /**
    * The distinguished root vertices, if known.
    */
-  private final Collection<IndexedWord> roots;
+  private final Set<IndexedWord> roots;
 
   private final DirectedMultiGraph<IndexedWord, SemanticGraphEdge> graph;
 
@@ -791,6 +791,10 @@ public Collection<IndexedWord> getRoots() {
     return roots;
   }
 
+  public boolean isRoot(IndexedWord vertex) {
+    return roots.contains(vertex);
+  }
+
   /**
    * Initially looks for nodes which have no incoming arcs. If there are any, it
    * returns a list of them. If not, it looks for nodes from which every other