使用 SQL 对二叉树节点进行分类

Question

I am trying to solve the problem mentioned at the below link

https://www.hackerrank.com/challenges/binary-search-tree-1/problem

I have written the following code.Please help me where i am going wrong

Select q.Node,case
              WHEN q.Node NOT IN q.Parent THEN 'Leaf' 
              WHEN q.Node IN q.Parent AND q.Node NOT IN q.Higher_Parent THEN 'Inner'
              WHEN q.Node IN q.Parent AND q.Node IN q.Higher_Parent THEN 'Root'
              END as NodeType
from (
SELECT B1.N as Node,B1.P as Parent,B2.P as Higher_Parent FROM 
BST B1 INNER JOIN BST B2 
ON B1.P = B2.N
ORDER BY Node ASC
) q

N P HP
1 2 5
2 5 NULL
3 2 5
6 8 5
8 5 NULL
9 8 5

Where should i modify the above code to work. Sure there are other concise codes for the same problem but please help me with this code for learning purpose.

Answer 1

select n.n,
  case 
    when max(p.n) is null then 'root'
    when max(c.n) is null then 'leaf'
    else 'inner'
  end as type
from bst n
left join bst p on p.n = n.p
left join bst c on c.p = n.n
group by n.n

Result:

| n |  type |
|---|-------|
| 1 |  leaf |
| 2 | inner |
| 3 |  leaf |
| 5 |  root |
| 6 |  leaf |
| 8 | inner |
| 9 |  leaf |

Demo: http://sqlfiddle.com/#!9/ba3c76/1

Answer 2

You need to find the child instead of the higher parent .

Select distinct c.n, 
        case when c.P is null then 'Root' 
             when b1.N is null then 'Leaf'
             else 'Inner' end
from BST c
left join BST b1
on c.N = b1.P
order by c.n

Answer 3

You can use if else to solve it as well:

SELECT B.N, IF(B.P IS NULL, 'Root', IF((SELECT COUNT(*) FROM BST AS A WHERE A.P=B.N)>0,'Inner','Leaf')) 
FROM BST AS B 
ORDER BY B.N;

Answer 4

You can use CASE statement as follows:

SELECT N, CASE WHEN P IS NULL THEN 'Root' 
WHEN(SELECT COUNT(*) FROM BST WHERE P = T.N) > 0 THEN 'Inner'
ELSE 'Leaf'
END
FROM BST T
ORDER BY N;

Answer 5

-- SQL SERVER working Solution

select N, case when p is null then "Root" when N in (select P from BST) THEN "Inner" Else "Leaf" end as "P" from BST ORDER BY N;

Answer 6

See the below code. It might be easy to comprehend.

select n, 'Leaf' as nodename from bst
where n not in (select distinct p from bst where p is not null)
union
select n, 'Root' as nodename from bst
where p is null
union
select n, 'Inner' as nodename from bst
where n in (select distinct p from bst)
and p is not null
order by n asc;