MySQL Solution for Tree Node Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe given SQL query categorizes nodes in a tree represented by the "tree" table into th
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe given SQL query categorizes nodes in a tree represented by the "tree" table into three types: \'Root\', \'Inner\', and \'Leaf\' based on their parent-child relationships. The query retrieves the ID of each node and assigns a corresponding type based on the conditions specified in the CASE statement.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. The main query starts with the SELECT keyword followed by the column names id and the CASE statement.\n2. The CASE statement consists of multiple WHEN conditions and an ELSE condition to determine the type of each node.\n3. In the first WHEN condition, p_id is null checks if the parent ID is null, indicating a root node. In this case, the type is set to \'Root\'.\n4. In the second WHEN condition, p_id is not null and id in (select distinct p_id from tree) checks if the node has a parent ID and if its ID appears in the set of distinct parent IDs in the "tree" table. This condition identifies inner nodes. In this case, the type is set to \'Inner\'.\n5. The ELSE condition is executed for nodes that do not match the previous conditions, which indicates leaf nodes. In this case, the type is set to \'Leaf\'.\n6. The query retrieves the results from the "tree" table.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this query depends on the size of the "tree" table and the execution efficiency of the subquery and the CASE statement. The subquery (select distinct p_id from tree) retrieves the distinct parent IDs, which may require scanning the "tree" table. The time complexity can vary but is typically linear or close to linear in the size of the "tree" table.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of the query depends on the number of nodes in the "tree" table and the storage required for the result set. The database engine needs to store the intermediate results of the subquery and the final result set containing the ID and type of each node. The space complexity is proportional to the number of nodes returned in the result set.\n\n# Code\n```\n# Write your MySQL query statement below\nselect id, \ncase\nwhen p_id is null then \'Root\'\nwhen p_id is not null and id in (select distinct p_id from tree) then \'Inner\'\nelse \'Leaf\' end as Type\nfrom tree\n```
['MySQL']