Return distinct items in asp-mvc4 list - asp.net

Question

How would I return a unique list 'listColl' I tried using 'Distinct9' at the bottom with 'return listColl;' but received an error.but it didnt work. Currently this return a duplicate list of items which populates a treeview coded in query.
public List<SPList> GetAllLibraries(string webURL)
{
var listColl = new List<SPList>();
ClientContext _ctx = new ClientContext(webURL);
try
{
var currentWeb = _ctx.Web;
var AllLists = currentWeb.Lists;
_ctx.Load(AllLists);
_ctx.ExecuteQuery();
var query = from list in currentWeb.Lists
where list.BaseType == BaseType.DocumentLibrary
select list;
var listCollection = _ctx.LoadQuery(query.Include(myList => myList.Title,
myList => myList.Id,
myList => myList.RootFolder.ServerRelativeUrl,
myList => myList.ParentWebUrl,
myList => myList.Hidden,
myList => myList.IsApplicationList));
_ctx.ExecuteQuery();
// /*
listColl.AddRange(from list in listCollection
where !list.Hidden
select new SPList
{
Title = list.Title,
ListGUID = list.Id.ToString(),
RootFolderServerRelativeUrl = list.RootFolder.ServerRelativeUrl,
ParentWebUrl = list.ParentWebUrl
});
// } */
foreach (var Item in listCollection)
{
listColl.Add(new SPList
{
Title = Item.Title,
RootFolderServerRelativeUrl = Item.RootFolder.ServerRelativeUrl,
ListGUID = Item.Id.ToString()
});
}
}
catch (Exception ex)
{
// error log
string error = ex.Message + " Error within GetAllLibraries ";
}
return listColl;
}

Answer 1

Can you do this?
return listColl.Distinct();

Answer 2

listColl.AddRange((from list in listCollection
where !list.Hidden
select new SPList
{
Title = list.Title,
ListGUID = list.Id.ToString(),
RootFolderServerRelativeUrl = list.RootFolder.ServerRelativeUrl,
ParentWebUrl = list.ParentWebUrl
}).Distinct());

Answer 3

A quick fix solved the problem in the mean time. The reason I can no do Distinct() is because of SPList.
List<SPList> distinct = new List<SPList>();
List<String> ids = new List<string>();
foreach (var lst in listColl)
{
if (!ids.Contains(lst.ListGUID))
{
distinct.Add(lst);
ids.Add(lst.ListGUID);
}
}
return distinct;