List构建树形结构

List构建树形结构

// 目标树形数据结构
public class LocationDto {

    /**
     * 空间id.
     */
    @Schema(description = "空间id", name = "id")
    private UUID id;

    /**
     * 子节点.
     */
    @Schema(description = "子节点", name = "children")
    private List<LocationDto> children;

    /**
     * 父节点id.
     */
    @Schema(description = "父节点id", name = "parentId")
    private UUID parentId;
}

// 源数据
public class LocationEntity {
    @TableId
    protected UUID id;
  
    /**
     * 父级id.
     */
    @TableField("parent_id")
    private UUID parentId;
}


方法实现：
     /**
     * 将List<LocationEntity>转成树形结构.
     * 次根节点的LocationEntity.parentId为null，没有最上层的根节点(或者说根节点可能为多个)
     *
     * @param searchDto 查询参数
     * @return 成功返回SUCCESS, 失败返回相应错误码
     */
public ApiResultDto<List<LocationDto>> getTreeByParams(LocationSearchDto searchDto, SecurityUser user) {
        List<LocationEntity> locationEntities = locationDao.findByParams(searchDto); // 数据源
        List<LocationDto> locationDtos = locationEntities.stream().map(t -> {
            LocationDto dto = new LocationDto();
            dto.setId(t.getId());
            dto.setName(t.getName());
            iterChild(dto); // 递归构建树形结构
            return dto;
        }).collect(Collectors.toList());
        return ApiResultDto.success(locationDtos);
    }

    /**
     * 遍历子节点.
     *
     * @param dto 空间dto
     */
    private void iterChild(LocationDto dto) {
        List<LocationEntity> locations = locationDao.findChildById(dto.getId());
        if (CollectionUtils.isEmpty(locations)) {
            return;
        }
        List<LocationDto> locationDtos = locations.stream().map(t -> {
            LocationDto childDto = new LocationDto();
            childDto.setId(t.getId());
            childDto.setName(t.getName());
            //遍历子节点
            iterChild(childDto);
            return childDto;
        }).collect(Collectors.toList());
        dto.setChildren(locationDtos);
    }