Is this code for getting the dependency tree of a file correct?

Question

function dep_tree(graph,node,prev_deps){
 return uniq(flatten(graph[node].map(function(child_node){ 
 if (contains(prev_deps,child_node))
 throw "Circular reference between "+child_node+" and "+node;
 return dep_tree(graph,child_node,prev_deps ? prev_deps.concat(node) : [node]) 
 })).concat(node));
};

• graph is an object such as {a:[],b:["a"],c:[],d:["b","c"]}, that links a file, ex: "d" to it's dependencies "b" "c".
• file is the the node to search.
• prev_deps is internal, to keep track of visited nodes

The function returns the full dependency tree of that node. For example, dep_tree(that_obj_above,"d") = ["b","c","a"]. Is this code correct, overall? Is it correctly checking for circular dependencies? Is there a name for what I'm doing?

• John Dvorak: uniq, flatten are imported from underscore.js?
• MaiaVictor: @JanDvorak Yes, I forgot to mention, sorry. Do I know you? That name sounds familiar.
• John Dvorak: We have met before
• MaiaVictor: @JanDvorak Oh sure you're the guy who made my fuzzy_match faster than assembly, thanks for that! Actually that was your only answer...!? Why!?
• John Dvorak: I don't come often to CR. I know I should.

Answer 1

Not a lot of code to review,

• To reduce confusion, I would use _.uniq() rather than point straight to uniq()
• lowerCamelCase is preferred, child_node -> childNode
• dep_tree is unfortunate, I like recursiveDependencies better

• I would have set prev_deps = prev_deps || []; prior to return, it would make your recursive call less golfic:

  return dep_tree(graph,child_node, prev_deps.concat(node) )