How do you solve Maximum Level Sum of a Binary Tree on LeetCode?

Maximum Level Sum of a Binary Tree (LeetCode #1161) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using BFS allows us to efficiently calculate level sums without redundant calculations.

What is the brute force solution for Maximum Level Sum of a Binary Tree?

The brute force approach for Maximum Level Sum of a Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. We can traverse the tree level by level, calculating the sum of values at each level. This approach is straightforward but inefficient as it recalculates sums for nodes multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Level Sum of a Binary Tree?

Maximum Level Sum of a Binary Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Breadth-First Search, Level Order Traversal.

What are the interview tips for Maximum Level Sum of a Binary Tree?

Always clarify the tree structure and traversal method before coding. Discuss your thought process aloud to demonstrate your understanding. Consider edge cases, such as trees with only one node or negative values.

What are common mistakes when solving Maximum Level Sum of a Binary Tree?

Not using a queue for level order traversal, leading to incorrect level sums. Forgetting to check for null children, which can cause errors in traversal.

#1161

Maximum Level Sum of a Binary Tree

Medium

TreeDepth-First SearchBreadth-First SearchBinary TreeBreadth-First SearchLevel Order Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a breadth-first search (BFS) approach to traverse the tree level by level while calculating the sum of node values at each level. This is efficient as we only traverse each node once.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue for BFS and add the root node.
2Step 2: For each level, calculate the sum of node values and track the maximum sum and corresponding level.
3Step 3: Return the level with the maximum sum.

solution.py30 lines

1# Full working Python code
2from collections import deque
3class TreeNode:
4    def __init__(self, val=0, left=None, right=None):
5        self.val = val
6        self.left = left
7        self.right = right
8
9def maxLevelSum(root):
10    if not root:
11        return 0
12    queue = deque([root])
13    max_sum = float('-inf')
14    level = 0
15    max_level = 0
16    while queue:
17        level += 1
18        level_sum = 0
19        for _ in range(len(queue)):
20            node = queue.popleft()
21            level_sum += node.val
22            if node.left:
23                queue.append(node.left)
24            if node.right:
25                queue.append(node.right)
26        if level_sum > max_sum:
27            max_sum = level_sum
28            max_level = level
29    return max_level
30

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the queue used for BFS, which can hold up to n nodes in the worst case.

1Using BFS allows us to efficiently calculate level sums without redundant calculations.
2Tracking the maximum sum and its level during traversal helps avoid a second pass.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.