使用 stream 流构建树(不使用递归)

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前言

最近遇到一个地区数据需要转换成树的需求，研究了一种新方法，不使用传统递归的方式，而是使用 stream 流的方式构建树。

代码实现

定义测试实体类

import lombok.Data;
import java.util.List;/*** <h2> </h2>** @author xymy* @date 2023/12/21 17:17*/
@Data
public class TsModel {private String id;private String name;private String pid;private List<TsModel> childs;
}

实现方法

public static void main(String[] args) {String str = "[\n" +"    {\"id\":\"1\",\"name\":\"张三\",\"pid\":\"\"},\n" +"    {\"id\":\"2\",\"name\":\"张三2\",\"pid\":\"\"},\n" +"    {\"id\":\"3\",\"name\":\"张三3\",\"pid\":\"\"},\n" +"    {\"id\":\"4\",\"name\":\"张三4\",\"pid\":\"\"},\n" +"    {\"id\":\"33\",\"name\":\"张三33\",\"pid\":\"3\"},\n" +"    {\"id\":\"330\",\"name\":\"张三330\",\"pid\":\"3\"},\n" +"    {\"id\":\"3301\",\"name\":\"张三330\",\"pid\":\"330\"},\n" +"    {\"id\":\"41\",\"name\":\"张三41\",\"pid\":\"4\"}\n" +"]";List<TsModel> tsModels = JSONArray.parseArray(str, TsModel.class);Map<String, List<TsModel>> pModels = tsModels.stream().collect(Collectors.groupingBy(ts -> StringUtils.isBlank(ts.getPid()) ? "" : ts.getPid()));List<TsModel> childTs = new ArrayList<>();for (TsModel tsModel : tsModels) {String id = tsModel.getId();List<TsModel> childModel = pModels.get(id);if (CollectionUtils.isNotEmpty(childModel)) {tsModel.setChilds(childModel);childTs.addAll(childModel);}}Set<String> childSeqs = childTs.stream().map(t -> t.getId()).collect(Collectors.toSet());List<TsModel> newTree = tsModels.stream().filter(tsm -> !childSeqs.contains(tsm.getId())).collect(Collectors.toList());System.out.println(JSON.toJSONString(newTree));}

打印结果：

[{"id": "1","name": "张三","pid": ""},{"id": "2","name": "张三2","pid": ""},{"childs": [{"id": "33","name": "张三33","pid": "3"},{"childs": [{"id": "3301","name": "张三330","pid": "330"}],"id": "330","name": "张三330","pid": "3"}],"id": "3","name": "张三3","pid": ""},{"childs": [{"id": "41","name": "张三41","pid": "4"}],"id": "4","name": "张三4","pid": ""}
]

至此，就实现了列表转成树的逻辑。