Started with a plain recursive DFS and it clicked pretty fast.
Start by clarifying the problem and edge cases, then present a recursive DFS solution that traverses the entire tree and sums nodes within the range. Discuss time and space complexity, and finally explain how the solution can be optimized for a BST by pruning branches that cannot contain valid nodes.
Pro tip: Mention that for a general binary tree, you must visit every node, but for a BST, you can prune subtrees where all values are outside the range, reducing time complexity to O(k) where k is the number of nodes in the range. This shows you understand the trade-offs and can optimize based on constraints.
Ask about input assumptions: Is the tree empty? Can low > high? Are node values unique? Should we consider inclusive bounds? This ensures you handle all cases correctly.
Write a function that recursively traverses left and right subtrees, adding the node's value if it falls within [low, high]. Base case: null node returns 0.
Time: O(n) since every node is visited. Space: O(h) for recursion stack, where h is tree height (O(n) worst case, O(log n) for balanced tree).
If the tree is a BST, prune branches: if node->val < low, skip left subtree; if node->val > high, skip right subtree. This reduces time to O(k) where k is nodes in range, but worst-case still O(n).
Compare general vs BST approach: BST pruning improves average case but not worst-case. Mention testing with empty tree, single node, all nodes in/out of range, and skewed trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
My instinct was to collect all values, sort, then relink nodes.
Clarify that the general case requires sorting the combined list, while the sorted case allows a linear merge. For the general case, either sort each list then merge, or concatenate and sort; for the sorted case, use a two-pointer merge. Present C++ code, analyze time and space complexity, and discuss trade-offs.
Pro tip: Mention that if the lists are already sorted, a two-pointer merge achieves O(n+m) time and O(1) extra space, which is optimal. For unsorted lists, emphasize that comparison-based sorting has an Ω((n+m) log(n+m)) lower bound, so any approach must be at least that efficient.
Ask whether the lists are sorted or unsorted, and confirm that nodes must be reused (no new nodes). Discuss edge cases like empty lists, single nodes, duplicates, and negative numbers.
Explain that the combined list must be sorted. Propose either sorting each list individually then merging, or concatenating and sorting the whole list. Mention that sorting a linked list can be done with merge sort in O(n log n) time and O(log n) space (due to recursion) or O(1) space with iterative merge sort.
If both lists are already sorted, use a two-pointer technique: maintain pointers to the heads of both lists, repeatedly attach the smaller node to the result list, and advance that pointer. Handle remaining nodes by linking the rest of the non-empty list.
Write clean C++ code for the sorted merge (and optionally for the general case). Use a dummy node to simplify list construction, and ensure node reuse by reassigning next pointers without allocating new nodes.
For the sorted case, time is O(n+m) and space is O(1). For the unsorted case, time is O((n+m) log(n+m)) and space depends on the sorting method (O(log n) for recursive merge sort, O(1) for iterative). Compare with alternative approaches like using an array.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.