binary tree

The diameter is not the tree's height, and the path does not have to pass through the root. The clean solution derives 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 divides the rest into left and right subtrees. The naive version 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 distinct from the answer you track: this problem stuffs three traps into one DFS. Here is the clean O(n) solution and the reasoning that holds up under follow-ups.