需要找到叶子节点、内部节点和根节点

Question

I am trying to solve this question which is on hacker rank

I got the desired output but not accepted by hacker rank

Input:

Output:

My trial 1:

with tb1 as
(
   select N,P from binary_tree
), tb2 as 
(
   select N,P from binary_tree
)
select t2.N as t2_N,
case 
     when t2.P is null and t1.P is null  and t1.N is null then 'root'
     when t2.P is not null and t1.P is null and t1.N is not null then 'inner'
     ELSE 'leaf'
end as RESULT
from tb2 t2 LEFT JOIN tb1 t1 ON t2.P = t1.N order by t2.N;

My trial 2:

with tb1 as
(
   select N,P from BST
), tb2 as 
(
   select P from BST
)
select distinct t.* from (select t1.N as tn,
case 
    when t1.N is not null and t2.P is not null and t1.P is null then 'root'
    when t1.N is not null and t2.P is not null and t1.P is not null then 'inner'
    when t1.N is not null and t2.P is null and t1.P is not null then 'leaf'
end as RESULT
from tb1 t1 LEFT JOIN tb2 t2 on t1.N = t2.P) t order by tn;

My above querys gives me desired output but not accepted

can some one figure it out why so?

please explain me how to solve this kind of binary tree questions and any trick to solve it

Answer 1

Finally after lots of trial able to write perfect query

with tb1 as
(
   select N,P from BST
), tb2 as 
(
   select P from BST
)
select distinct t.* from (select t1.N as tn,
case 
    when t1.N is not null and t2.P is not null and t1.P is null then 'Root'
    when t1.N is not null and t2.P is not null and t1.P is not null then 'Inner'
    when t1.N is not null and t2.P is null and t1.P is not null then 'Leaf'
end as RESULT
from tb1 t1 LEFT JOIN tb2 t2 on t1.N = t2.P) t order by tn;

Please if you have any other method to solve it share it

Answer 2

I assume this is meant to test your understanding of Oracle "connect by" queries, rather than standard joins. Roots can be identified by null parent, leafs are detected by the connect_by_isleaf pseudocolumn, and all other nodes are "inner". So, the query can be written like this:

select  n,
        case when p is null             then 'Root'
             when connect_by_isleaf = 1 then 'Leaf'
             else                            'Inner' end as result
from    binary_tree
start   with p is null
connect by p = prior n
order   by n  --  if needed
;