求救一个sql语句或者存储过程

[happyct]

表记录：

1002
100201
1002018523532
100201435X46413
100202
1002026335332
2001
200101
200102


我现在需要得到哪些节点是分支节点，哪些是叶子节点。实际上：1002，100201，100202，2001是分支节点，其他都是叶子节点。可以肯定的是，同个父节点下的分支节点（非叶子节点）的长度是一样的。

[happyct]

如果可能的话，最好知道每个分支节点的层次。

[CT_LXL]

引用 1 楼 happyct 的回复:

如果可能的话，最好知道每个分支节点的层次。

1--非叶子 0-叶子节点 lvl为层次：

with t as
 (select '1002' c1
    from dual
  union all
  select '100201' c1
    from dual
  union all
  select '1002018523532' c1
    from dual
  union all
  select '100201435X46413' c1
    from dual
  union all
  select '100202' c1
    from dual
  union all
  select '1002026335332' c1
    from dual
  union all
  select '2001' c1
    from dual
  union all
  select '200101' c1
    from dual
  union all
  select '200102' c1
    from dual)
select t3.c1, t3.lvl, t4.is_leaf
  from (select t2.c1, count(1) lvl
          from t t1, t t2
         where t2.c1 like t1.c1 || '%'
         group by t2.c1
         order by t2.c1) t3,
       (select c1,
               decode((select count(*)
                        from t t2
                       where t2.c1 like t1.c1 || '%'
                         and t2.c1 <> t1.c1),
                      0,
                      0,
                      1) is_leaf
          from t t1) t4
 where t3.c1 = t4.c1;

[一颗向往数据的心丶]

引用 2 楼 zlloct 的回复:

[quote=引用 1 楼 happyct 的回复:] 如果可能的话，最好知道每个分支节点的层次。

1--非叶子 0-叶子节点 lvl为层次：

with t as
 (select '1002' c1
    from dual
  union all
  select '100201' c1
    from dual
  union all
  select '1002018523532' c1
    from dual
  union all
  select '100201435X46413' c1
    from dual
  union all
  select '100202' c1
    from dual
  union all
  select '1002026335332' c1
    from dual
  union all
  select '2001' c1
    from dual
  union all
  select '200101' c1
    from dual
  union all
  select '200102' c1
    from dual)
select t3.c1, t3.lvl, t4.is_leaf
  from (select t2.c1, count(1) lvl
          from t t1, t t2
         where t2.c1 like t1.c1 || '%'
         group by t2.c1
         order by t2.c1) t3,
       (select c1,
               decode((select count(*)
                        from t t2
                       where t2.c1 like t1.c1 || '%'
                         and t2.c1 <> t1.c1),
                      0,
                      0,
                      1) is_leaf
          from t t1) t4
 where t3.c1 = t4.c1;

[/quote] ++

[happyct]

非常感谢，万分感谢。