Java 如果给定层次 struct 级别，如何从其预序穿越构造n元树

我收到了一个基于SON格式的响应，其中n元树的PRENER穿越以JSONArray给出.我需要将其转换为树 struct 并发送回来作为响应.

给出回应

{
  "ControllingArea" : "TZUS",
  "TopNodes" : false,
  "Language" : "EN",
  "TONODES" : [
    {
      "Groupname" : "INC",
      "Hierlevel" : 0,
      "Valcount" : 0,
      "Descript" : "INC"
    },
    {
      "Groupname" : "INC-BBSR",
      "Hierlevel" : 1,
      "Valcount" : 0,
      "Descript" : "INC1"
    },
    {
      "Groupname" : "INC-STPI",
      "Hierlevel" : 2,
      "Valcount" : 0,
      "Descript" : "INC2"
    },
    {
      "Groupname" : "INC-FORT",
      "Hierlevel" : 2,
      "Valcount" : 0,
      "Descript" : "INC3"
    },
    {
      "Groupname" : "INC-BBL",
      "Hierlevel" : 1,
      "Valcount" : 0,
      "Descript" : "INC4"
    },
    {
      "Groupname" : "INC-PUNE",
      "Hierlevel" : 1,
      "Valcount" : 0,
      "Descript" : "INC5"
    }
  ],
  "TOVALUE" : [
    {
      "Valfrom" : "",
      "Valto" : ""
    }
  ]
}

这里的TONODES代表树上的Preorder穿越，也在对象中给出了它所属的级别.

树应该看起来像(UI视图).

tree image

我想发送的回复应该是:

{
    "INC": {
        "Groupname": "INC",
        "Hierlevel": 0,
        "Valcount": 0,
        "Descript": "INC",
        "INC-BBSR": {
            "Groupname": "INC-BBSR",
            "Hierlevel": 1,
            "Valcount": 0,
            "Descript": "INC1",
            "INC-STPI": {
                "Groupname": "INC-STPI",
                "Hierlevel": 2,
                "Valcount": 0,
                "Descript": "INC2"
            },
            "INC-FORT": {
                "Groupname": "INC-FORT",
                "Hierlevel": 2,
                "Valcount": 0,
                "Descript": "INC3"
            }
        },
        "INC-BBL": {
            "Groupname": "INC-BBL",
            "Hierlevel": 1,
            "Valcount": 0,
            "Descript": "INC4"
        },
        "INC-PUNE": {
            "Groupname": "INC-PUNE",
            "Hierlevel": 1,
            "Valcount": 0,
            "Descript": "INC5"
        }
    }
}

我try 在网上搜索了很多生成N-Ary树的方法，但没有得到任何满意的结果.因为解决方案说，如果你知道n，u就能够做出树，但我不知道这里的n，因为它可以是随机的.

我还try 在对象的更高层次 struct 即将到来时使用堆栈将 node 插入到其中，并在对象的较低层次 struct 即将到来时弹出.

@Override
    public ResponseDTO getTreeStructure(JSONObject sapResponse) {

        JSONObject response = new JSONObject();

        Stack<JSONObject> treeStack = new Stack<>();

        JSONArray iteratingArray = sapResponse.getJSONArray("TONODES");

        for(int i = 0 ; i < iteratingArray.length(); i++){
            JSONObject tempObject= iteratingArray.getJSONObject(i);

            if(treeStack.empty()) {
                treeStack.push(tempObject);
                response.put(tempObject.getString("Groupname"), tempObject);
            }else if(treeStack.peek().getInt("Hierlevel") < tempObject.getInt("Hierlevel")){
                response.getJSONObject(treeStack.peek().getString("Groupname")).put(tempObject.getString("Groupname"), tempObject);
                treeStack.push(tempObject);
            }else if(treeStack.peek().getInt("Hierlevel") >= tempObject.getInt("Hierlevel")){
                while(!treeStack.empty() && treeStack.peek().getInt("Hierlevel") >= tempObject.getInt("Hierlevel")){
                    treeStack.pop();
                }
                response.getJSONObject(treeStack.peek().getString("Groupname")).put(tempObject.getString("Groupname"), tempObject);
                treeStack.push(tempObject);
            }
        }

        return new ResponseDTO(200, "ok", response);

        //return null;
    }

问题是我无法让之前的对象推入响应.我无法追踪父 node 来获取并到达当前 node 进行推送.

0(N0) -> 1(N1) -> 2(N2)

response.getJSONObject(N0).getJSONObjectI(N1).put(N2.getString("Groupname"), N2);

但相反，在我的方法中，我得到了

response.getJSONObject(N1).put(N2)

I can get the node 1 from the top of the stack, but I cannot track down the parent 0th node to push the 2nd object under 1st.

如果以某种方式可以按照层次 struct 获得之前的 node ，我就能够创建树.

请更正我的代码或告诉我解决这个问题的更好方法.