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

byAkshay Daga (APDaga) -
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HackerRank: [Advanced Select - 4/5] Binary Tree Nodes | CASE, WHEN, IF, Sub-query in SQL
I started studying SQL from a very famous site - HackerRank. Here I will try to provide multiple approaches & solutions to the same problem. It will help you learn and understand SQL in a better way.

Please make use of my blog posts for learning purpose only and feel free to ask your questions in the comment box below in case of any doubt.

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Recommended SQL Courses:
  1. Udemy: SQL - MySQL for Data Analytics and Business Intelligence
  2. LinkedIn: Master SQL for Data Science
  3. edX: Databases: Advanced Topics in SQL
  4. edureka: SQL Essentials Training & Certification
  5. Coursera: Learn SQL Basics for Data Science
  6. Eduonix: SQL Queries

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 BST table is described as follows:

BST_Columns

Sample Input:
The BST 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: 
  1. Parent for Root Node is Null.
  2. Any node which is the parent node of other nodes is an Inner node
  3. All remaining nodes are Leaf nodes. 


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: 
  1. Parent for Root Node is Null.
  2. Any node which is the parent node of other nodes is an Inner node
  3. All remaining nodes are Leaf nodes. 


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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-Akshay P Daga
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