Count nodes in complete tree
Given a complete binary tree encoded as a dense level-order array (every level full except possibly the last, which is filled left to right with no gaps), return the number of nodes in O(log^2 n) time. Do not just return the array length: implement the height comparison: if a node's leftmost path height equals its rightmost path height, its subtree is perfect with 2^h - 1 nodes; otherwise recurse on both children. Return the node count.
Implement
count_complete_nodes(level: list[int]) → intExamples
in
[[1,2,3,4,5,6]]out6What 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
[[1,2,3,4,5,6]]6not run yetsample