1161. Maximum Level Sum of a Binary Tree

Reference to LeetCode is here, solution is here and visualization is here!

Thoughts and Talk

Problem Statement Simplification

Hints to Approach the Problem

1. Tree Traversal: Choose a traversal method. Depth-first (DFS) or breadth-first (BFS) are options, but BFS is particularly useful for processing nodes level by level.

2. Sum Tracking: Maintain a running total of node values at each tree level.

3. Level Identification: Use a mechanism to keep track of which tree level is being processed, which is crucial for correct sum accumulation.

4. Comparison and Reset: At each new tree level, compare the accumulated sum with the maximum recorded sum and reset the sum for the next level.

Approaches Explored

Approach 1: Using Lists to Store Level Sums

• Pros: Lists provide a direct mapping of levels to sums, which is intuitive and easy to implement.

• Cons: This method can be space-intensive, particularly for unbalanced trees, and slightly inefficient if the tree is very deep.

Approach 2: Using Recursive DFS to Compute Sums

• Pros: Recursion simplifies the traversal logic, and the method is inherently elegant.

• Cons: It can be less efficient in terms of space due to recursive calls, and handling level tracking is more complex with recursion.

Approach 3: BFS with a Queue (Implemented in the Provided Code)

• Pros: Directly corresponds to the needs of the problem by facilitating real-time processing and comparison of sums at each level.

• Cons: Slightly more complex queue management, but the complexity is manageable and generally offers better performance compared to other methods.

Finalized Approach

Executable Code

TheTechCruise.com Pyodide Terminal

from collections import deque
from typing import Optional, List

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

def create_tree(values: List[Optional[int]]) -> Optional[TreeNode]:
    if not values:
        return None
    root = TreeNode(values[0])
    queue = deque([root])
    i = 1
    while queue and i < len(values):
        node = queue.popleft()
        if values[i] is not None:
            node.left = TreeNode(values[i])
            queue.append(node.left)
        i += 1
        if i < len(values) and values[i] is not None:
            node.right = TreeNode(values[i])
            queue.append(node.right)
        i += 1
    return root

class Solution:
    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0

        queue = deque([(root, 1)])
        max_sum = float('-inf')
        max_level = 0
        current_level = 1
        current_sum = 0

        while queue:
            node, level = queue.popleft()

            if level != current_level:
                if current_sum > max_sum:
                    max_sum = current_sum
                    max_level = current_level
                current_sum = 0
                current_level = level

            current_sum += node.val

            if node.left:
                queue.append((node.left, level + 1))
            if node.right:
                queue.append((node.right, level + 1))

        if current_sum > max_sum:
            max_sum = current_sum
            max_level = current_level

        return max_level

# Example tree creation and execution
values = [-100, -200, -300, -20, -5, -10, 7]
root = create_tree(values)
solution = Solution()
result = solution.maxLevelSum(root)
print("The level with the maximum sum is:", result)

Code Explanation

1. Queue Initialization: The root node and its level are initially enqueued, setting the stage for level-by-level processing.

2. Loop through Queue: Nodes are processed sequentially, with checks to ensure that sums are accumulated and compared at correct levels.

3. Level Change Detection and Sum Comparison: The sum of each level is compared against the maximum found so far. This detection triggers at the transition between levels.

4. Queue Operations for Child Nodes: Child nodes are enqueued with incremented level values, ensuring the BFS traversal continues correctly.

5. Final Comparison: The final comparison post-loop ensures that the sum of the last level is also considered for maximum sum calculation.

Visualization

Complexity Discussion

• Time Complexity: O(N), where N is the number of nodes in the tree. Each node is processed exactly once, making the approach very efficient.

• Space Complexity: O(M), where M is the maximum width of the tree at any level. This could potentially involve storing many nodes in the queue at once, especially in the case of a full binary tree.