Java 8 中的 Stream 轻松遍历树形结构
直接上代码
测试实体类
/*** Menu** @author lcry*/
@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;}}
核心代码：
@Test
public void testtree(){//模拟从数据库查询出来，公众号Java精选，有惊喜！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;
}
打印效果
