我正在尝试反序列化一个json，如下所示

{
    "timestamp": 123456789,
    "ownerAssets" : [
        {
            "ownerId" : 123,
            "location" : "USA",
            "assets" : [
                {
                    "car" : true,
                    "make" : "honda",
                    "model" : "crv"
                },
                {
                    "fruit" : true,
                    "name" : "apple",
                    "sweetness" : "high"
                    "count": 5
                }
            ]
        },
        {
            "ownerId" : 456,
            "location" : "USA",
            "assets" : [
                {
                    "car" : true,
                    "make" : "toyota",
                    "model" : "highlander"
                },
                {
                    "fruit" : true,
                    "name" : "orange",
                    "sweetness" : "low",
                    "count": 5
                }
            ]
        }
    ]
}

我如何设计DTO模型来解析它呢？我希望避免使用协议，而更喜欢使用具有关联值的枚举。
我认为以下几点是显而易见的：

struct StuffDTO: Codable {
    let timestamp: Int
    let ownerAssets: [OwnerAsset]
}

extension StuffDTO {
    struct OwnerAsset: Codable {
        let ownerId: Int
        let location: String
        let assets: [Asset]
    }
}

我可以定义一个uber对象，它可以将所有内容和if-then-else解析为不同的枚举值（见下文），这些枚举值具有不同类型的关联值

struct Asset: Codable {
    // car fields
    let car: Bool?
    let make: String?
    let model: String?
    
    // fruit fields
    let fruit: Bool?
    let name: String?
    let sweetness: String?
    let count: Int?
}

但是我想知道是否有一种聪明的方法可以直接将json解析成枚举值，而不需要解析成一个uber对象或者不使用多态性/协议。

enum Asset: Codable {
    case car(CarDTO)
    case fruit(FruitDTO)

    init() {
        // what would this custom json parsing logic look like?
    }
}

struct CarDTO: Codable {
    let make: String
    let model: String
}

struct FruitDTO: Codable {
    let name: String
    let sweetness: String
    let count: Int
}