Arcesium Maximum Width Level

Learning Outcome

Turn the Arcesium Maximum Width Level interview variant into a clear brute-force baseline, optimized pattern, and implementation plan.

Original Interview Statement

Given a binary tree, return the maximum width between leftmost and rightmost non-null nodes at any level, counting null gaps.

Examples

Brute Force Approach

Store every null placeholder in BFS. This can explode exponentially for deep sparse trees.

Optimized Approach

Assign heap-style indices to non-null nodes. Width of a level is lastIndex - firstIndex + 1. Normalize each level to avoid overflow.

Exact Pseudocode

queue root with index 0
for each level:
  width = last - first + 1
  pop nodes and push children with 2*i and 2*i+1 using normalized i
return max width
  

Reference Code

from collections import deque

def width_of_binary_tree(root):
    if not root:
        return 0
    q = deque([(root, 0)])
    best = 0
    while q:
        first = q[0][1]
        last = q[-1][1]
        best = max(best, last - first + 1)
        for _ in range(len(q)):
            node, idx = q.popleft()
            idx -= first
            if node.left:
                q.append((node.left, 2 * idx))
            if node.right:
                q.append((node.right, 2 * idx + 1))
    return best
  

Complexity

Edge Cases

• Sparse deep tree
• Single node
• Index overflow

Follow-ups

• Return the level number too
• Use DFS with first index per depth

Nearest Practice References

• Maximum Width of Binary Tree

Common Mistakes

• Copying the nearest LeetCode solution without checking the changed rule.
• Skipping duplicate or boundary cases.
• Not stating the brute force before the optimized approach.