← Bytedance Interview Insights
I had failed this exact problem at a previous ByteDance interview and somehow got it again here.
Use level-order traversal (BFS) while assigning each node a position index as if the tree were a complete binary tree. For each level, compute the width as the difference between the rightmost and leftmost indices plus one, and track the maximum.
Pro tip: Mention that indices can grow exponentially, so use 64-bit integers (long) to avoid overflow, and note that the problem can also be solved with DFS by tracking the leftmost index per level.
Confirm that width counts null nodes between the leftmost and rightmost non-null nodes at each level, and that the answer is the maximum over all levels.
Decide between BFS (level-order) and DFS. BFS is more intuitive for level-by-level processing, while DFS can be more space-efficient.
For each node, assign an index: root gets 0, left child gets 2*i, right child gets 2*i+1. This simulates a complete binary tree and allows width calculation.
During BFS, for each level, record the first and last indices. Width = last - first + 1. Update the global maximum.
Consider empty tree (return 0), single node (return 1), and skewed trees. Analyze time O(N) and space O(N) for BFS, or O(H) for DFS.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.