Node Values Between Given Levels in Binary Tree

Unit 9 Session 2 (Click for link to problem statements)

Problem Highlights

💡 Difficulty: Medium
⏰ Time to complete: 30+ mins
🛠️ Topics: Tree, Breadth-First Search, Queue, Level Order Traversal

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?

Can the tree be empty?
- Yes, if the tree is empty, return an empty list.
What should be returned if the tree has only one node and the levels are within the range?
- Return a list with that single node's value.

HAPPY CASE

    3 (root)
   / \
  9   20
     /  \
    15   7

Input: root, start_level = 2, end_level = 3
Output: [9, 20, 15, 7]
Explanation: The nodes between level 2 and 3 are [9, 20, 15, 7].

EDGE CASE
 
    None (root)

Input: root, start_level = 1, end_level = 2
Output: []
Explanation: The tree is empty, so return an empty list.

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 problems, we want to consider the following approaches:

Breadth-First Search (BFS): Useful for level order traversal while tracking the levels.
Queue: Used to manage the order of node exploration.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a BFS approach to traverse the tree level by level. Use a queue to keep track of nodes to be explored along with their corresponding levels, and a list to store the node values within the specified levels.

1) If the tree is empty, return an empty list.
2) Create a list to hold the node values within the specified levels.
3) Create an empty queue and add the root node with level 1.
4) While the queue is not empty:
    a) Pop the next node and its level from the queue.
    b) If the node's level is within the specified range, add its value to the list.
    c) If the current level is less than the end level, enqueue the left child with level + 1.
    d) If the current level is less than the end level, enqueue the right child with level + 1.
5) Return the list of node values within the specified levels.

⚠️ Common Mistakes

Forgetting to handle the case where the tree is empty.
Not correctly managing the queue for BFS.
Incorrectly identifying and adding the levels and their corresponding values.

4: I-mplement

Implement the code to solve the algorithm.

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def get_level_range(root, start_level, end_level):
    if not root:
        return []
    
    result = []
    queue = deque() 
    queue.append((root, 1)) # Each element in the queue is a tuple (node, current_level)
    
    while queue:
        node, level = queue.popleft()
        
        if start_level <= level <= end_level:
            result.append(node.val)
        
        if level < end_level:
            if node.left:
                queue.append((node.left, level + 1))
            if node.right:
                queue.append((node.right, level + 1))
    
    return result

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 an input to check for the expected output
Catch possible edge cases and off-by-one errors
For a starting point, try checking your code against the Happy/Edge Case(s) above

6: E-valuate

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

Time Complexity:

O(N), where N is the number of nodes in the tree.
This is because we must visit each node exactly once to gather the node values.

Space Complexity:

O(N), where N is the number of nodes in the tree.
This is because the queue and result list could hold up to N nodes in the worst case (when the tree is completely unbalanced).

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