Stream流的方式递归出树形结构
遇到构建菜单,构建树形结构,数据库一般就使用父id来表示,为了降低数据库的查询压力,我们可以使用Java8中的Stream流一次性把数据查出来,然后通过流式处理
实体类
@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<Menu> 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<Menu> 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<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 */ private 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; }