How to: JSON Serialize a node/list instead of a field

Question

I have my class for serialize variables, but items and fields should be nodes not fields, I need to get this output:

{
  "items": [
    {
      "email": "string",
      "fields": [
        {
          "name": "string",
          "value": "string",
        }
      ]
    }
  ]
}

This is the class i have but I dont know how to implement items and fields nodes:

public class DopplerJSONSerializer {

    public String name;
    public String email;
    public String items;
    public String fields;

    public String upsertSuscribersJSON(String suscriberEmail, String suscName) {
        DopplerJSONSerializer suscWrapper = new DopplerJSONSerializer();
        suscWrapper.email = suscriberEmail;
        suscWrapper.name = suscName;
        String listJSON = JSON.serializePretty(suscWrapper, true);
        return listJSON;
    }
}

Since right now its returning:

{"name":"New Susc", "email":"asd@asd.com"}

Answer 1

I used JsonToApex. This is a free tool to perform such tasks.

For that structure class would be like:

public class DopplerJSONSerializer1{
    public cls_items[] items;
    class cls_items {
        public String email;    //string
        public cls_fields[] fields;
    }
    class cls_fields {
        public String name; //string
        public String value;    //string
    }
    public static DopplerJSONSerializer1 parse(String json){
        return (DopplerJSONSerializer1) System.JSON.deserialize(json, DopplerJSONSerializer1.class);
    }

    static testMethod void testParse() {
        String json=        '{'+
        '  "items": ['+
        '    {'+
        '      "email": "string",'+
        '      "fields": ['+
        '        {'+
        '          "name": "string",'+
        '          "value": "string",'+
        '        }'+
        '      ]'+
        '    }'+
        '  ]'+
        '}';
        DopplerJSONSerializer1 obj = parse(json);
        System.assert(obj != null);
    }
}

To Serialize:

DopplerJSONSerializer1 wrapper1 = new DopplerJSONSerializer1();
DopplerJSONSerializer1.cls_items items = new DopplerJSONSerializer1.cls_items();

DopplerJSONSerializer1.cls_fields fields = new DopplerJSONSerializer1.cls_fields();
fields.name = 'name';
fields.value = 'value';

items.email = 'me@you.com';
items.fields = new List<DopplerJSONSerializer1.cls_fields>{ fields };

wrapper1.items = items;

JSON.serializePretty(wrapper1, true);