Morris inorder traversal
Given a binary tree encoded as a level-order list (None marks a missing child; children of None are omitted), return its inorder traversal as a list of node values. Implement Morris traversal (threaded binary tree) using O(1) extra space beyond the output — do NOT use a stack or recursion. The first element is the root; an empty list or [None] means an empty tree.
Implement
morris_inorder(level: list) → list[int]Examples
in
[[1,null,2,3]]out[1,3,2]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 30 min
solution.py
InputExpectedGot
[[1,null,2,3]][1,3,2]not run yetsample