Minimum Depth of Secret Path

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Unit 9 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20 mins
• 🛠️ Topics: Binary Trees, Recursion, Depth-First Search, Breadth-First Search

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 should be returned if the door tree is None?
  • Return 0 since there are no paths in the tree.
• What if the tree has only one node?
  • Return 1 since the root node is also the leaf node, making the minimum depth 1.
• Is the tree guaranteed to be balanced?
  • The problem assumes the input tree is balanced when calculating time complexity.

HAPPY CASE
Input: door = Room("Door", Room("Attic"), Room("Cursed Room", Room("Crypt"), Room("Haunted Cellar")))
Output: 2
Explanation: The shortest path is from "Door" -> "Attic".

Input: door = Room("Door")
Output: 1
Explanation: The tree has only one node, so the minimum depth is 1.

EDGE CASE
Input: door = None
Output: 0
Explanation: The tree is empty, so return 0.

Input: door = Room("Door", Room("Attic"), None)
Output: 2
Explanation: The shortest path is from "Door" -> "Attic".

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 problems involving finding the minimum depth of a binary tree, we can consider the following approaches:

• Depth-First Search (DFS): Use DFS to explore each path from the root to the leaves and keep track of the minimum depth encountered.
• Breadth-First Search (BFS): Use BFS to explore the tree level by level and return the depth of the first leaf node encountered.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

If you used a BFS approach, check out Minimum Depth of Secret Path II.

Plan (DFS Approach)

1) Base Case:

• If the door tree is None, return 0.
• If the current node is a leaf (no children), return 1. 2) Recursive Check:
• Recursively calculate the minimum depth of the left subtree.
• Recursively calculate the minimum depth of the right subtree.
• Return the minimum of the two depths plus one for the current node.

DFS Implementation

Pseudocode:

1) Define the base cases:
   * If `door` is `None`, return `0`.
   * If the current node is a leaf, return `1`.

2) Recursively calculate the depth of the left subtree (`left_depth = min_depth(door.left)`).

3) Recursively calculate the depth of the right subtree (`right_depth = min_depth(door.right)`).

4) Return `1 + min(left_depth, right_depth)` as the minimum depth.

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 min_depth(door):
    if not door:
        return 0
    
    if not door.left and not door.right:
        return 1
    
    if not door.left:
        return 1 + min_depth(door.right)
    
    if not door.right:
        return 1 + min_depth(door.left)
    
    return 1 + min(min_depth(door.left), min_depth(door.right))

5: R-eview

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

• Trace through your code with the input door = Room("Door", Room("Attic"), Room("Cursed Room", Room("Crypt"), Room("Haunted Cellar"))):
  • The DFS should correctly identify the shortest path as "Door" -> "Attic" with a minimum depth of 2.

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 tree.

• Time Complexity: O(N) because each node in the tree must be visited once.
• Space Complexity: O(N) due to the recursive call stack in the worst case, assuming a balanced tree.