当JSON中缺少且不可为空的属性时，如何抛出JsonSerializationException？

using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
using System;

public class RequiredPropertyConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return true;
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        JObject jsonObject = JObject.Load(reader);

        // 检查是否缺少属性
        if (jsonObject["requiredProperty"] == null)
        {
            throw new JsonSerializationException("缺少必需的属性：requiredProperty");
        }

        // 反序列化对象
        var targetObject = Activator.CreateInstance(objectType);
        serializer.Populate(jsonObject.CreateReader(), targetObject);
        return targetObject;
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

using Newtonsoft.Json;

// 创建自定义的JsonSerializerSettings
var settings = new JsonSerializerSettings();
settings.Converters.Add(new RequiredPropertyConverter());

// 序列化对象
var json = JsonConvert.SerializeObject(yourObject, settings);

// 反序列化对象
var deserializedObject = JsonConvert.DeserializeObject<YourObjectType>(json, settings);