用jackson解析时如何防止多余的字符串转换步骤？

Question

我有一些代码可以将安静的响应转换为递归结构。但是，我想解析成树的内容被包装到“treeprop”属性中。有没有更方便的方式解析真实内容？

ObjectMapper mapper = new ObjectMapper();
JsonFactory jfac = new JsonFactory();
JsonParser jp = jfac.createParser(inputStream);
JsonNode rootNode = mapper.readTree(jp);

JsonNode path = rootNode.path("treeprop");
String realContent = mapper.writeValueAsString(path);

MyTree mt = mapper.readValue(realContent, MyTree.class);
inputStream.close();

请注意，解析本身不是这里的问题。上面的代码确实将 inputStream 正确地转换为树。但是，json 通常大于 1MB，因此将其临时存储在 String 中不能提高运行时效率。

{
  "treeprop": {
    "id": 0,
    "children": [
      {
        "id": 2,
        "children": [
          {
            "id": 3,
            "children": []
          },
          {
            "id": 4,
            "children": []
          }
        ]
      },
      {
        "id": 1,
        "children": []
      }
    ]
  }
}

这个类看起来大致是这样的

class MyTree {
   public Integer id;
   public List<MyTree> children;
}

所以真正的问题是：是否有更有效的替代方案来实现相同的效果：

JsonNode path = rootNode.path("treeprop");
String realContent = mapper.writeValueAsString(path);
MyTree mt = mapper.readValue(realContent, MyTree.class);

Answer 1

public class MyTreeWrapper {
    private MyTree treeprop;

    // getter, setter
}

...

ObjectMapper mapper = new ObjectMapper();
MyTreeWrapper wrapper = mapper.readValue(inputStream, MyTreeWrapper.class);
MyTree tree = wrapper.getTreeprop();

替代方案：

MyTree tree = mapper.reader(MyTree.class).readValue(rootNode.path("treeprop"));