How to ignore property if it is null while creating JSON response? [duplicate]

I have 1 class like below:

public class Tables
    {
        public string Name { get; set; }
        public string[] Columns { get; set; }
    }

string[] selectedTables = { "Table1", "Table2"};
using (var conection = new SqlConnection("MyconnectionString"))
{
    connection.Open();

    var tables = (
        from table in connection.GetSchema("Tables").AsEnumerable()
        let name = (string)table["TABLE_NAME"]
        where selectedTables.Contains(name)
        let catalog = (string)table["TABLE_CATALOG"]
        let schema = (string)table["TABLE_SCHEMA"]
        select new Tables // this should really be called Table
        {
            Name = name,
            Columns = (
                from column in connection.GetSchema("Columns", new [] { catalog, schema, name }).AsEnumerable()
                select (string)column["COLUMN_NAME"]).ToArray()
        }).ToList();

    return tables;
}

Here i am creating response to return data to service:

     var response = tables.
                      Select
                   (
                        t => new
                          {
                            id = t.Id,
                            tables = t.Tables
                            .Select
                            (
                              x => new
                               {
                                   name = x.Name,
                                   columns = x.Columns 
                               }
                             ).ToList()
                          }
                    ).ToList();

  return Json(response);

Below is my JSON:

Now when generating columns data for each tables if columns is null for suppose Table1 then I don't want this columns field to be present in my JSON.

Is this possible because I follow this Question in which it is written that it is not possible.

Update: I am using asp.net mvc so I am doing like this to create json - is that possible?

return Json(response);

This is not duplicate question as I am using JSON class of MVC to generate JSON result.