Unit 8 Session 1 (Click for link to problem statements)
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?
HAPPY CASE
Input: TreeNode('AND', TreeNode(True), TreeNode(False))
Output: False
Explanation: The 'AND' operation between True and False yields False.
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 involves evaluating a simple Boolean expression represented in a binary tree format, where the root node defines the operation (AND/OR) and the children provide the operands.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Use the value of the root node to determine which Boolean operation to perform on the values of its children.
1) Retrieve the Boolean values from the children nodes.
2) Perform the operation specified by the root's value on the children's values.
3) Return the result of the operation.
Implement the code to solve the algorithm.
def tree_expression(root):
"
Evaluates a boolean operation ('AND' or 'OR') stored in the root node on its two child nodes.
"
# Retrieve the boolean values from the children
left_value = root.left.val
right_value = root.right.val
# Perform the operation based on the root's value
if root.val == 'AND':
return left_value and right_value
elif root.val == 'OR':
return left_value or right_value
else:
return False
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(1)
as the function only checks and processes three nodes regardless of any other factors.O(1)
because no extra space is used beyond the input nodes.