树形结构，你会怎么处理

最近闲来无事，想起以前在开发需求的时候，遇到了返回属性菜单的需求，当时用了递归算法去实现的，但是总觉得写的不好，容易造成栈溢出。现在想想为啥不利用jdk8的新特性来实现，下面我就直接放出代码，有需要的自己参考一下：

import java.util.List;

@Data
@NoArgsConstructor
@AllArgsConstructor
public class Menu {

    private Integer id;
    private String name;
    private Integer parentId;
    private List<Menu> childList;

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

import com.alibaba.fastjson.JSON;

import java.util.Arrays;
import java.util.List;
import java.util.Objects;
import java.util.stream.Collectors;

public class Demo {

    public static void main(String[] args){
        //模拟从数据库查询出来
        List<Menu> 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<Menu> 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
     */
    public static List<Menu> getChildrens(Menu root, List<Menu> all) {
        List<Menu> 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;
    }
}