这个怎么样（根据你的样本数据和结果）。如果你的数据很大，也许有更好的解决方案。而且你没有指定如果30也是20的父：

with steps as (
    select 0 step, cast(null as varchar2(10)) auto_skip from dual
    union ALL
    select 10,null from dual
    union all
    select 20, 'skip' from dual
    union all
    select 21,'skip' from dual
    union all 
    select 30 ,null from dual
)
select step as child,LEAD (step,1) OVER (ORDER BY step) AS parent from STEPS
union
select step as child,LEAD (step,1) OVER (ORDER BY step) AS parent from STEPS where auto_skip is null;

CHILD PARENT 
----- ------ 
    0     10 
   10     20 
   20     21 
   21     30 
   30        
   10     30 

试试这个解决方案，用UNION两种不同的方法来计算父节点，第二种方法是递归的，必须区分几次跳跃或只有一次跳跃。它可以在测试数据上工作，但应该在更多的数据上检查。

with data(step,auto_skip) as (
    select 0, null from dual union all
    select 10, null  from dual union all
    select 20, 'skip1' from dual union all
    select 21, 'skip1' from dual union all
    select 22, 'skip2' from dual union all
    select 30, null  from dual -- union all
),
pdata as (
    select d.* from (
        select d.step, d.auto_skip, 
            lead(d.step,1) over(order by d.step) as natural_father,
            lead(d.auto_skip,1) over(order by d.step) as natural_father_skip
        from data d
    ) d
),
rdata as (
    select level as lvl, connect_by_isleaf as isleaf, connect_by_root(step) as rstep, d.* 
    from pdata d
    start with natural_father_skip <> nvl(auto_skip,'*')
    connect by 
        prior natural_father = step and prior natural_father_skip = auto_skip
)
select * from (
    select step as child, natural_father as parent from pdata
    union all
    select rstep, case when lvl > 2 then step else natural_father end from rdata
    where isleaf = 1 
)
order by child, parent
;

0   10
10  20
10  22
20  21
21  22
21  30
22  30
30