start with connect by level 实现树
具体的sql为
select level treelevel,
       eva.state,
       t.id,
       case
         when t.parentid is null then
          ''
         else
          t.parentid || ''
       end parentid,
       t.summary,
       t.score,
       t.content,
       t.standardscore,
       t.selfAssessmentScore,
       t.remark,
       t.scoreReason,
       t.ruleType,
       case
         when (select count(1) from abc dt where dt.parentid = t.id) = 0 then
          '1'
         else
          '0'
       end leaf,
       case
         when level = 3 then
          '0'
         else
          '1'
       end expanded
  from abc t, def eva
where t.defid = eva.id
   and t.defid = #value#
start with t.id in (select id
                       from abc d
                      where d.parentid is null
                        and d.defid = #value#)
connect by t.parentid = prior t.id ORDER SIBLINGS BY t.seq
****************************************************************
这段sql实现的是一个树形表，使用start with connect by prior level来实现，这个树有两张表：abc和def，abc是def的子表(abc.defid = def.id)。def存的是这个树叫什么名字，是哪个单位，哪个月份的树，abc存的是这个树具体的细则。
leaf和expanded映射成javaBean后是两个boolean类型的值，用来判断图标。
因此，排除def的干扰后，由abc的递归查询来实现这颗树：
start with t.id in (select id
                       from abc d
                      where d.parentid is null
                        and d.defid = #value#)
connect by t.parentid = prior t.id ORDER SIBLINGS BY t.seq

首先确定递归查询的范围：select id
                       from abc d
                      where d.parentid is null
                     &n