Felt pretty comfortable with the base case but the follow-up threw me off.
Use a depth-first search (DFS) to traverse the tree, passing the current number formed so far. At each leaf, add the current number to a running sum. This approach is efficient and easy to implement recursively.
Pro tip: Clarify edge cases upfront, such as an empty tree or a tree with a single node, and discuss how to handle integer overflow if the tree is deep. This shows attention to detail and robustness.
Confirm that each root-to-leaf path forms a number by concatenating digits, and we need the sum of all such numbers. Ask clarifying questions about tree size, digit range, and expected output type.
Decide between recursive DFS, iterative DFS with a stack, or BFS. DFS is natural because it maintains the path from root to current node.
Define a helper function that takes a node and the current number formed so far. At each node, update the number as current * 10 + node.val. If it's a leaf, return the number; otherwise, return the sum of recursive calls on left and right children.
Check for empty tree (return 0) and single node (return its value). Consider potential integer overflow and discuss using long or modular arithmetic if needed.
Time complexity is O(N) where N is number of nodes, as each node is visited once. Space complexity is O(H) for recursion stack, where H is tree height, which is O(N) in worst case.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.