How do you solve Tree Node on LeetCode?

Tree Node (LeetCode #608) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of trees is crucial for identifying node types.

What is the brute force solution for Tree Node?

The brute force approach for Tree Node is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each node to see if it has children by searching through the entire table. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Tree Node?

Tree Node uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Tree Node?

Always clarify the structure of the data before diving into the solution. Think about edge cases, such as a single node or a completely unbalanced tree. Practice writing SQL queries as they often come up in data-related interviews.

What are common mistakes when solving Tree Node?

Failing to account for the case when there's only one node. Not considering the possibility of multiple child nodes for inner nodes.

#608

Tree Node

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By using a single pass to create a set of child nodes, we can efficiently classify each node in one go.

⚙️

Algorithm

3 steps

1Step 1: Create a set of all child node ids by selecting distinct p_id values from the Tree table.
2Step 2: For each node, check if it is in the child set to determine if it is a leaf or inner node.
3Step 3: Classify the node as 'Root', 'Inner', or 'Leaf' based on its p_id and presence in the child set.

solution.py10 lines

1WITH ChildNodes AS (
2    SELECT DISTINCT p_id FROM Tree WHERE p_id IS NOT NULL
3)
4SELECT id,
5       CASE
6           WHEN p_id IS NULL THEN 'Root'
7           WHEN id IN (SELECT * FROM ChildNodes) THEN 'Inner'
8           ELSE 'Leaf'
9       END AS type
10FROM Tree;

ℹ

Complexity note: The time complexity is O(n) because we only need to scan through the nodes a couple of times: once to create the child set and once to classify each node.

1Understanding the structure of trees is crucial for identifying node types.
2Using sets can significantly improve lookup times when classifying nodes.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.