### Thoughts and Talk

When approaching tree-related problems, it's crucial to select the right traversal strategy to align with the problem's requirements. This problem is a classic example where understanding the structure and leveraging level-based processing can significantly simplify the solution. Similar strategies can be applied in systems design where hierarchical data processing is necessary, proving that these techniques are not only academically interesting but also practically valuable in technology and beyond. I would assume that it is also useful in analysis of program run times and profilers, HTML DOM parsers and optimizers amongst other things — but yet to be verified

### Problem Statement Simplification

The problem involves finding the level in a binary tree that has the maximum sum of its node values. Essentially, this is similar to determining which floor in a skyscraper has the most occupants, or which department in a company has the highest sales in a given quarter. Understanding this helps in resource allocation and optimization in real-world scenarios.

### Hints to Approach the Problem

**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.

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

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

**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

The BFS approach using a queue was selected because it efficiently meets the requirement of processing data level by level. This method allows for immediate updates and comparisons of sums, making it highly suitable for real-time applications, such as network traffic monitoring or organizing data in web servers.

### 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

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

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

**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.

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

**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.

With that we can conclude on how to calculate the max level sum of the given binary tree!