需求场景

有下面一张区域表，典型的树形结构设计。
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现前端需要后端返回树形数据结构用于构造展示树。
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代码实战

实验数据构造：
首先我们根据数据库结构创建实体对象

  1. /**
  2.  * 区域平台
  3.  * @author:Jam
  4.  */
  5. @Data
  6. public class Platform {
  7.     private String id;
  8.     private String parentId;
  9.     private String name;
  10.     private String platformCode;
  11.     private List<Platform> children;
  13.     public Platform(String id, String platformCode,String parentId, String name) {
  14.         this.id = id;
  15.         this.parentId = parentId;
  16.         this.name = name;
  17.         this.platformCode = platformCode;
  18.     }
  19. }

为了便于演示我们就不连接数据库，而是直接使用Junit5的@BeforeEach注解初始化一份结构数据。

  1. public class PlatformTest {
  3.     private final List<Platform> platformList = Lists.newArrayList();
  4.     private ObjectMapper objectMapper = new ObjectMapper();
  6.     @BeforeEach
  7.     private void init(){
  8.         Platform platform0 = new Platform("1","001","0","集团");
  9.         Platform platform1 = new Platform("2","QYPT001","1","销委会");
  10.         Platform platform2 = new Platform("3","QYPT002","2","吉龙大区");
  11.         Platform platform3 = new Platform("4","QYPT003","2","江苏大区");
  12.         Platform platform4 = new Platform("5","QYPT004","4","南京分区");
  14.         Platform platform5 = new Platform("6","QYPT005","1","教育BG");
  15.         Platform platform6 = new Platform("7","QYPT006","6","华南大区");
  16.         Platform platform7 = new Platform("8","QYPT007","6","华东大区");
  18.         platformList.add(platform0);
  19.         platformList.add(platform1);
  20.         platformList.add(platform2);
  21.         platformList.add(platform3);
  22.         platformList.add(platform4);
  23.         platformList.add(platform5);
  24.         platformList.add(platform6);
  25.         platformList.add(platform7);
  26.     }
  27. }

一般方法：

  1. 首先查到根节点，parent_id = 0
  2. 通过根节点id获取到所有一级节点，parent_id = 1

  3. 递归获取所有节点的子节点，然后调用setChildren()方法组装数据结构。

    双重循环：

    通过双重循环确定父子节点的关系。 ```java @SneakyThrows @Test public void test1(){ System.out.println(platformList.size()); List result = Lists.newArrayList(); for (Platform platform : platformList) {

    //获取根节点 if(platform.getParentId().equals(“0”)){ result.add(platform); }

    for(Platform child : platformList){ if(child.getParentId().equals(platform.getId())){

    1.  platform.addChild(child);

    } } }

    System.out.println(objectMapper.writeValueAsString(result)); }

public void addChild(Platform platform){ if(children == null){ children = new ArrayList<>(); } children.add(platform); }

  1. <a name="ckThb"></a>
  2. ### 双重遍历：
  3. 第一次遍历借助hashmap存储父节点与子节点的关系，第二次遍历设置子节点，由于map中已经维护好了对应关系所以只需要从map取即可。
  4. ```java
  5. @SneakyThrows
  6. @Test
  7. public void test2(){
  8.   Map<String, List<Platform>> platformMap = new HashMap<>();
  10.   platformList.forEach(platform -> {
  11.     List<Platform> children = platformMap.getOrDefault(platform.getParentId(), new ArrayList<>());
  12.     children.add(platform);
  13.     platformMap.put(platform.getParentId(),children);
  14.   });
  16.   platformList.forEach(platform -> platform.setChildren(platformMap.get(platform.getId())));
  18.   List<Platform> result = platformList.stream().filter(v -> v.getParentId().equals("0")).collect(Collectors.toList());
  20.   System.out.println(objectMapper.writeValueAsString(result));
  22. }

Stream 分组：

  1. @SneakyThrows
  2. @Test
  3. public void test4(){
  4.   Map<String, List<Platform>> groupMap = platformList.stream().collect(Collectors.groupingBy(Platform::getParentId));
  5.   platformList.forEach(platform -> platform.setChildren(groupMap.get(platform.getId())));
  6.   List<Platform> collect = platformList.stream()
  7.     .filter(platform -> platform.getParentId().equals("0")).collect(Collectors.toList());
  8.   System.out.println(objectMapper.writeValueAsString(collect));
  9. }

此处主要通过Collectors.groupingBy(Platform::getParentId)方法对platformList按照parentId进行分组，分组后父节点相同的都放一起了。
然后再循环platformList，给其设置children属性。
执行完成后已经形成了多颗树，最后我们再通过filter()方法挑选出根节点的那颗树即可。