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;
Answers:
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Method 1
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)
Method 2
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).
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0