Use a depth-first search (DFS) traversal, passing the current number formed so far (as an integer) down the recursion. At each leaf, add the current number to a running total. Return the total sum.
Pro tip: Clarify edge cases upfront: what if the tree is empty? What if a node has only one child? Also, mention that you can avoid integer overflow by using modulo or by noting constraints, but typically the sum fits in a 32-bit integer for reasonable tree depths.
Restate the problem: each root-to-leaf path forms a number by concatenating digits. We need the sum of all such numbers. Confirm with the interviewer if the tree can be empty or if digits are 0-9.
Decide between DFS (recursive or iterative) and BFS. DFS is natural because we need to carry the current number down the path. Mention that BFS would require storing the current number with each node in the queue.
Write a helper function that takes a node and the current number formed so far. At each node, update current number as current * 10 + node.val. If leaf, add to sum; else recurse on children.
If node is null, return 0. If leaf, return the current number. Otherwise, return sum of left and right subtrees. This naturally accumulates the total.
Time complexity: O(N) where N is number of nodes, as each node is visited once. Space complexity: O(H) for recursion stack, where H is tree height (worst case O(N) for skewed tree).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The array being pre-sorted is the hint, but my first instinct was still to just iterate through.
Use binary search to find the first and last occurrence of the target, then return the range of indices. This achieves O(log n) time by performing two modified binary searches instead of scanning linearly.
Pro tip: Clarify with the interviewer whether the output should be a list of indices or a range (start, end) — this shows attention to detail and can simplify the solution. Also, mention edge cases like empty array or target not present.
Confirm the expected output format (list of indices vs. range) and discuss edge cases such as empty array, target absent, or all elements equal to target.
Modify binary search to find the leftmost index of the target by continuing to search left even when the target is found.
Similarly, modify binary search to find the rightmost index by continuing to search right when the target is found.
If both searches return valid indices, return the range or list of indices; otherwise, return an empty list or indicate absence.
State that time complexity is O(log n) and space is O(1) (or O(k) for output). Walk through a few test cases to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.