binary tree

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.