飘乙己：List转Tree有4种写法！

Posted 2022-03-01 飘渺Jam

孔乙己说：回字有四种写法。

飘乙己也说，list转tree也有4种写法，你用的是哪种？

需求场景

有下面一张区域表，典型的树形结构设计。

现前端需要后端返回树形数据结构用于构造展示树。

本篇文章我们就来介绍一下在这种场景下后端构建树形数据结构，也就是通过list转tree的4种写法。

代码实战

1. 首先我们根据数据库结构创建实体对象

/**
 * 区域平台
 * @author:Jam
 */
@Data
public class Platform 
    private String id;
    private String parentId;
    private String name;
    private String platformCode;
    private List<Platform> children;

    public Platform(String id, String platformCode,String parentId, String name) 
        this.id = id;
        this.parentId = parentId;
        this.name = name;
        this.platformCode = platformCode;
    

2. 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。

public class PlatformTest 

    private final List<Platform> platformList = Lists.newArrayList();
    private ObjectMapper objectMapper = new ObjectMapper();

    @BeforeEach
    private void init()
        Platform platform0 = new Platform("1","001","0","集团");
        Platform platform1 = new Platform("2","QYPT001","1","销委会");
        Platform platform2 = new Platform("3","QYPT002","2","吉龙大区");
        Platform platform3 = new Platform("4","QYPT003","2","江苏大区");
        Platform platform4 = new Platform("5","QYPT004","4","南京分区");

        Platform platform5 = new Platform("6","QYPT005","1","教育BG");
        Platform platform6 = new Platform("7","QYPT006","6","华南大区");
        Platform platform7 = new Platform("8","QYPT007","6","华东大区");

        platformList.add(platform0);
        platformList.add(platform1);
        platformList.add(platform2);
        platformList.add(platform3);
        platformList.add(platform4);
        platformList.add(platform5);
        platformList.add(platform6);
        platformList.add(platform7);
    

最无节操的写法

这种写法毫无节操可言，全部通过数据库递归查询。

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。

这种写法我就不展示了，辣眼睛。都2021年了我见过不止一次在项目中出现这种写法。

双重循环

这种写法比较简单，也是比较容易想到的。通过双重循环确定父子节点的关系。

@SneakyThrows
@Test
public void test1()
  System.out.println(platformList.size());
  List<Platform> 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()))
        platform.addChild(child);
      
    
  

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

同时需要给Platform添加一个addChild()的方法。

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

双重遍历

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。

@SneakyThrows
@Test
public void test2()
  Map<String, List<Platform>> platformMap = new HashMap<>();

  platformList.forEach(platform -> 
    List<Platform> children = platformMap.getOrDefault(platform.getParentId(), new ArrayList<>());
    children.add(platform);
    platformMap.put(platform.getParentId(),children);
  );

  platformList.forEach(platform -> platform.setChildren(platformMap.get(platform.getId())));

  List<Platform> result = platformList.stream().filter(v -> v.getParentId().equals("0")).collect(Collectors.toList());

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

Stream 分组

@SneakyThrows
@Test
public void test4()
  Map<String, List<Platform>> groupMap = platformList.stream().collect(Collectors.groupingBy(Platform::getParentId));
  platformList.forEach(platform -> platform.setChildren(groupMap.get(platform.getId())));
  List<Platform> collect = platformList.stream()
    .filter(platform -> platform.getParentId().equals("0")).collect(Collectors.toList());
  System.out.println(objectMapper.writeValueAsString(collect));

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。