I got the BFS approach pretty quickly, track the index alongside each node, and at each level just subtract the leftmost index from the rightmost and add one.
Use BFS level-order traversal, assigning each node an index as in a complete binary tree (root=1, left=2i, right=2i+1). For each level, compute width as rightmost index minus leftmost index plus one, and track the maximum. To avoid integer overflow, use unsigned 64-bit integers or compute indices relative to the leftmost node of each level.
Pro tip: Mention that using relative indices per level (subtracting the leftmost index) prevents overflow even for very deep trees, and that this approach is essentially the same as LeetCode 662. Also, note that DFS with a depth parameter can achieve O(n) time and O(h) space, which might be more memory-efficient for skewed trees.
Confirm that width is defined by the span between leftmost and rightmost non-null nodes at each level using complete binary tree indexing. Discuss potential index overflow for deep trees (e.g., depth > 64) and the need for O(n) time.
Decide between BFS (level-order) and DFS (pre-order with depth). BFS naturally processes level by level, while DFS can save space. Both can achieve O(n) time.
For each node, assign an index: root=1, left=2i, right=2i+1. For each level, track the first and last index encountered. Width = last - first + 1. Update global maximum.
Use 64-bit unsigned integers for indices, or compute indices relative to the leftmost node of the current level to keep numbers small. Explain why this prevents overflow.
Time: O(n) since each node visited once. Space: O(n) for BFS queue or O(h) for DFS recursion stack. Discuss edge cases: empty tree, single node, skewed tree, complete tree.
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