I am trying to solve this problem,
You are given a table, BST, containing two columns: N and P, where N represents the value of a node in Binary Tree, and P is the parent of N.
Write a query to find the node type of Binary Tree ordered by the value of the node. Output one of the following for each node:
Root: If node is root node. Leaf: If node is leaf node. Inner: If node is neither root nor leaf node.
Input:
Desired Output:
1 Leaf
2 Inner
3 Leaf
5 Root
6 Leaf
8 Inner
9 Leaf
This is my query, can anyone tell me why it's not working?
select case
when P is NULL then CONCAT_WS(" ", N, 'Root')
when N not in (SELECT DISTINCT P FROM BST) then CONCAT_WS(" ", N, 'Leaf')
else CONCAT_WS(" ", N, 'Inner')
end
from BST ORDER BY N ASC;
CodePudding user response:
You have NULL
inside the P
column in which case an expression such as:
1 NOT IN (NULL, 2, 8, 5)
will return unknown
instead of the "expected" result true
(ref).
The solution is to make a subtle change like so:
N NOT IN (SELECT P FROM BST WHERE P IS NOT NULL)
Or use an EXISTS
query:
NOT EXISTS (SELECT * FROM BST AS bst2 WHERE bst2.P = bst.N)
CodePudding user response:
SELECT n,
ELT((1 (2 * (t1.p IS NULL)) (EXISTS (SELECT NULL FROM BST t2 WHERE t1.n=t2.p))), 'Leaf', 'Inner', 'Single', 'Root') type
FROM BST t1
ORDER BY n;
https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=287b1de2b2bb532d73619c19bcf8a86b
I add one more option 'Single' - the case when the node is root and leaf at the same time (node 4 in my fiddle).