# HackerRank: [SQL 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.

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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 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 PLACEFROM BST BORDER 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 PLACEFROM BST BORDER BY N;`

### Sample Output:

`1 Leaf2 Inner3 Leaf4 Inner5 Leaf6 Inner7 Leaf8 Leaf9 Inner10 Leaf11 Inner12 Leaf13 Inner14 Leaf15 Root`

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-Akshay P Daga