Bottom view of binary tree
Given a binary tree as a level-order array `arr` (null for missing), return its bottom view: for each horizontal distance (root at 0, left child -1, right child +1), the value of the node that would be seen from directly below. When several nodes share a horizontal distance, the one visited LAST in a left-to-right level-order (BFS) traversal wins. Return values ordered by horizontal distance, left to right. `0 <= node count <= 10000`.
Implement
bottom_view(arr: list[int]) → list[int]Examples
in
[[20,8,22,5,3,null,25]]out[5,8,3,22,25]What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 25 min
solution.py
InputExpectedGot
[[20,8,22,5,3,null,25]][5,8,3,22,25]not run yetsample