Java 8 中的 Stream 轻松遍历树形结构，是真的牛逼。。。

/**
 * Menu
 *
 * @author lcry
 * @date 2020/06/01 20:36
 */
    @Data
    @Builder
    public class Menu {
    /**
     * id
     */
     public Integer id;
     /**
     * 名称
     */
     public String name;
     /**
     * 父id ，根节点为0
     */
     public Integer parentId;
     /**
     * 子节点信息
     */
     public List childList;


    public Menu(Integer id, String name, Integer parentId) {
        this.id = id;
        this.name = name;
        this.parentId = parentId;
    }

    public Menu(Integer id, String name, Integer parentId, List childList) {
        this.id = id;
        this.name = name;
        this.parentId = parentId;
        this.childList = childList;
    }

}

递归组装树形结构：

@Test
public void testtree(){
    //模拟从数据库查询出来
    List menus = Arrays.asList(
            new Menu(1,"根节点",0),
            new Menu(2,"子节点1",1),
            new Menu(3,"子节点1.1",2),
            new Menu(4,"子节点1.2",2),
            new Menu(5,"根节点1.3",2),
            new Menu(6,"根节点2",1),
            new Menu(7,"根节点2.1",6),
            new Menu(8,"根节点2.2",6),
            new Menu(9,"根节点2.2.1",7),
            new Menu(10,"根节点2.2.2",7),
            new Menu(11,"根节点3",1),
            new Menu(12,"根节点3.1",11)
    );

    //获取父节点
    List collect = menus.stream().filter(m -> m.getParentId() == 0).map(
            (m) -> {
                m.setChildList(getChildrens(m, menus));
                return m;
            }
    ).collect(Collectors.toList());
    System.out.println("-------转json输出结果-------");
    System.out.println(JSON.toJSON(collect));
}

/**
 * 递归查询子节点
 * @param root  根节点
 * @param all   所有节点
 * @return 根节点信息
 */
private List getChildrens(Menu root, List all) {
    List children = all.stream().filter(m -> {
        return Objects.equals(m.getParentId(), root.getId());
    }).map(
            (m) -> {
                m.setChildList(getChildrens(m, all));
                return m;
            }
    ).collect(Collectors.toList());
    return children;
}