3 Node Product II

Unit 8 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 5 mins
• 🛠️ Topics: Trees, Binary Trees, Math Operations

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Question: What if one or both children nodes are missing?
  • Answer: If either child node is missing, return False because the product comparison cannot be completed.

HAPPY CASE
Input: TreeNode(6, TreeNode(2), TreeNode(3))
Output: True
Explanation: The product of the values of the two children (2 * 3) equals the value of the root (6).

EDGE CASE
Input: TreeNode(6, TreeNode(2), None)
Output: False
Explanation: The tree is missing one child node, hence the product comparison cannot be completed.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem is a typical validation problem within the tree structure that checks if a node’s value corresponds to a specific relation with its children's values (in this case, a product relation).

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Check if the root's value equals the product of its children’s values.

1) Check if root and both children are not None.
2) Calculate the product of the children's values.
3) Compare the product with the root's value.
4) Return True if they match, else return False.

⚠️ Common Mistakes

• Incorrect handling of cases where the tree has missing children, which would lead to incorrect calculations or exceptions.

4: I-mplement

Implement the code to solve the algorithm.

def check_tree(root):
    if not root or not root.left or not root.right:
        return False
    return root.val == root.left.val * root.right.val

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Test with various configurations of the tree to ensure correctness, especially where children nodes might be missing.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(1) as the operation only involves a direct comparison after a single multiplication.
• Space Complexity: O(1) because no additional space is used beyond the existing tree nodes.