trees and traversal

The diameter is not the tree's height, and the path need not pass through the root. The clean solution computes height and best path in one DFS, updating a global max at every node. Here is the answer and the subtlety candidates miss.

Preorder names the root; inorder splits the rest into left and right subtrees. Doing it naively is O(n squared); the strong answer uses a value-to-index map and a moving preorder pointer for O(n). Here is the answer and why both orders are required.

Negative subtrees, a path that bends through any node, and a return value that differs from the answer you track: this problem packs three traps into one DFS. Here is the clean O(n) solution and the reasoning that survives follow-ups.