Sum Inventory

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

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Binary Tree, Tree Traversal, Recursion

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?

• What does each node in the binary tree represent?
  • Each node represents the current stock of a flower variety in the store.
• How should the function behave if the tree is empty?
  • The function should return 0 if the tree is empty.

HAPPY CASE
Input: Binary tree with nodes [45, 12, 10, 20, 1, 15]
Output: 106
Explanation: The sum of all nodes is 45 + 12 + 10 + 20 + 1 + 15 = 106.

EDGE CASE
Input: Binary tree with only one node [50]
Output: 50
Explanation: The tree has only the root, so the sum is 50.

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.

For Tree Summation problems, we want to consider the following approaches:

• Binary Tree Traversal: Traverse the tree to calculate the sum of all nodes.
• Recursion: Use recursion to sum the values of the nodes by summing the values of each subtree.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the tree recursively, summing the values of each node by adding the sum of the left subtree and the right subtree.

1) If the current node is None, return 0.
2) Recursively calculate the sum of the left subtree.
3) Recursively calculate the sum of the right subtree.
4) The sum of the current node is its value plus the sum of the left and right subtrees.
5) Return the sum of the tree rooted at the current node.

⚠️ Common Mistakes

• Not correctly handling the base case where the tree is empty.
• Forgetting to include the current node's value in the total sum.

4: I-mplement

Implement the code to solve the algorithm.

class TreeNode:
    def __init__(self, value, left=None, right=None):
        self.val = value
        self.left = left
        self.right = right

def sum_inventory(root):
    if root is None:
        return 0
    
    # Recursive case: calculate the sum of left and right subtrees
    left_sum = sum_inventory(root.left)
    right_sum = sum_inventory(root.right)
    
    # The sum of the tree rooted at the current node
    return root.val + left_sum + right_sum

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 the examples given:
  • Input 1: Binary tree with nodes [45, 12, 10, 20, 1, 15]
  • Expected Output: 106
  • Input 2: Binary tree with only one node [50]
  • Expected Output: 50
  • Verify that the outputs match the expected results.

6: E-valuate

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

Assume N represents the number of nodes in the binary tree.

• Time Complexity: O(N) because the algorithm visits each node once.
• Space Complexity: O(H) where H is the height of the tree, due to the recursive call stack.