飘乙己：List转Tree有4种写法！

た入场券 2022-09-11 04:26 155阅读 0赞

> 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。

![image-20210625110242489][]

现前端需要后端返回树形数据结构用于构造展示树。

![image-20210628124838191][]

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;
        }
    
    }

1.  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。

[image-20210625110242489]: /images/20220828/67946662fa9f4483bc85f63a2865ecf3.png
[image-20210628124838191]: /images/20220828/966af13c2c1a40229cf1d0e98140ebb3.png