refactor 608

Author: fishercoder1534

Diff for: database/_608.sql

@@ -1,45 +1,3 @@
---608. Tree Node
---Given a table tree, id is identifier of the tree node and p_id is its parent node's id.
---
---+----+------+
---| id | p_id |
---+----+------+
---| 1  | null |
---| 2  | 1    |
---| 3  | 1    |
---| 4  | 2    |
---| 5  | 2    |
---+----+------+
---Each node in the tree can be one of three types:
---Leaf: if the node is a leaf node.
---Root: if the node is the root of the tree.
---Inner: If the node is neither a leaf node nor a root node.
---Write a query to print the node id and the type of the node. Sort your output by the node id. The result for the above sample is:
---+----+------+
---| id | Type |
---+----+------+
---| 1  | Root |
---| 2  | Inner|
---| 3  | Leaf |
---| 4  | Leaf |
---| 5  | Leaf |
---+----+------+
---Explanation
---
---Node '1' is root node, because its parent node is NULL and it has child node '2' and '3'.
---Node '2' is inner node, because it has parent node '1' and child node '4' and '5'.
---Node '3', '4' and '5' is Leaf node, because they have parent node and they don't have child node.
---
---And here is the image of the sample tree as below:
---			  1
---			/   \
---                      2       3
---                    /   \
---                  4       5
---Note
---
---If there is only one node on the tree, you only need to output its root attributes.
-
 --credit: https://leetcode.com/articles/tree-node/#approach-i-using-union-accepted
 select id, 'Root' as Type from tree where p_id is null
 union