[开发测试]java 两种方式生成递归菜单，支持树断裂情况下生成新的树

1、菜单类

@Data
public class Menu {
    private Long id;
    private String name;
    private Long pid;
    private List<Menu> children;

    public Menu(Long id, String name, Long pid) {
        this.id = id;
        this.name = name;
        this.pid = pid;
    }
}

2、父id确定情况下生成新的树

public static void test1(){
        List<Menu> menuList=new ArrayList<>();
        menuList.add(new Menu(1L,"菜单1",0L));
        menuList.add(new Menu(2L,"菜单2",1L));
        menuList.add(new Menu(3L,"菜单3",2L));
        menuList.add(new Menu(4L,"菜单4",5L));

        List<Menu> menus=menuList.stream().filter(m->m.getPid()==0).map(m->{
            m.setChildren(getChildrens(m,menuList));
            return m;
        }).collect(Collectors.toList());
        System.out.println(JSONUtil.toJsonPrettyStr(menus));
    }

3、递归生成子树?

private static List<Menu> getChildrens(Menu root, List<Menu> all) {
        List<Menu> children = all.stream().filter(m -> Objects.equals(m.getPid(), root.getId())).map(
                (m) -> {
                    m.setChildren(getChildrens(m, all));
                    return m;
                }
        ).collect(Collectors.toList());
        return children;
    }

4、有查询条件树断裂情况下根据id和pid生成新的树

public static void test2(){
        List<Menu> menuList=new ArrayList<>();
        menuList.add(new Menu(1L,"菜单1",0L));
        menuList.add(new Menu(2L,"菜单2",1L));
        menuList.add(new Menu(3L,"菜单3",2L));
        menuList.add(new Menu(4L,"菜单4",5L));

        List<Long> IdList=menuList.stream().map(s->s.getId()).collect(Collectors.toList());
        List<Long> PidList=menuList.stream().map(s->s.getPid()).collect(Collectors.toList());
        /**
         * 求两者的交集
         */
        List<Long> jiaoJiIdList=new ArrayList<>();
        for(Long id:IdList){
            if(PidList.contains(id)){
                jiaoJiIdList.add(id);
            }
        }

        /**
         * 过滤时所有pid在交集里面的数据，若pid在交集中则说明该元素有父类
         */
        List<Menu> menus=menuList.stream().filter(m->!jiaoJiIdList.contains(m.getPid())).map(m->{
            m.setChildren(getChildrens(m,menuList));
            return m;
        }).collect(Collectors.toList());
        System.out.println(JSONUtil.toJsonPrettyStr(menus));
    }