CODEWITHSUNDEEP
mysqlsql
DJ-coding
DJ-coding

Reputation: 323

Finding "Leaf" in a Tree using MySQL

I am trying to identify the "Leaf" in a tree but I am confused why my query is not giving me what I want.

Here is the Problem:

enter image description here

So my idea is that as long as the id is not in p_id column, then it should be a "Leaf"

select id, 'Leaf' as type
from Tree
where id not in (select distinct(p_id)
                 from Tree)

However, the query above is returning nothing to me.

The solution is pretty much the same as mine except it specify that the p_id cannot be NULL, then it is returning what I want.

select id
from Tree
where id not in(select distinct(p_id) 
                from Tree 
                where p_id IS NOT NULL)

I am confused as why does adding the where clause will make a difference?

Upvotes: 2

Views: 300

Answers (2)

Revathi N
Revathi N

Reputation: 1

select distinct a_id, status from( select a.id as a_id,a.pid as a_pid, b.id as b_id, b.pid as b_pid, case when b.id is not null and a.pid is not null then 'Inner' When b.id is null then 'leaf' when a.pid is null then 'root' else null end as 'status'

from leaf a left join leaf b on a.id = b.pid)temp

Upvotes: 0

slaakso
slaakso

Reputation: 9070

You guessed it. It is because NULL does not compare to anything. A value is not different from null and a value is not same as null.

You can get your resut with following query:

select distinct t.id, 
  if (p.id is null, 'Root', if (d.id is null, 'Leaf', 'Inner'))
from Tree t
  left join Tree p on p.id=t.p_id
  left join Tree d on d.p_id=t.id;

See dbfiddle.

Upvotes: 5

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