HackerRank: [SQL Advanced Select] (4/5) BINARY TREE NODES | case, when, if, sub-query in SQL

HackerRank: [Advanced Select - 4/5] Binary Tree Nodes |  CASE, WHEN, IF, Sub-query in SQL
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SQL Problem Statement:

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 Format:

The BSTĀ table is described as follows:

BST_Columns

Sample Input:
TheĀ BSTĀ table contains the following records:
BST_Sample_data

Sample Output:
1 Leaf 
2 Inner 
3 Leaf 
5 Root 
6 Leaf 
8 Inner 
9 Leaf


Explanation:
The Binary Tree below illustrates the sample:

Solution-1: Using CASE WHEN Statement (MySQL Query):

SELECT N,
CASE
   WHEN P IS NULL THEN 'Root'
   WHEN (SELECT COUNT(*) FROM BST WHERE B.N=P)>0 THEN 'Inner'
   ELSE 'Leaf'
END AS PLACE
FROM BST B
ORDER BY N;

NOTE:Ā 
  1. ParentĀ for Root Node is Null.
  2. Any node which is the parent node of other nodes is anĀ Inner node
  3. All remaining nodes are Leaf nodes.Ā 



Solution-2: Using IF Statement (MySQL Query):

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

NOTE:Ā 
  1. ParentĀ forĀ RootĀ Node isĀ Null.
  2. Any node which is the parent node of other nodes is anĀ InnerĀ node
  3. All remaining nodes areĀ LeafĀ nodes.Ā 



Sample Output:

1 Leaf
2 Inner
3 Leaf
4 Inner
5 Leaf
6 Inner
7 Leaf
8 Leaf
9 Inner
10 Leaf
11 Inner
12 Leaf
13 Inner
14 Leaf
15 Root

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