I went with DFS and accumulated the current number by multiplying by 10 and adding the node value as I went down.
Use a depth-first search (DFS) to traverse the tree, maintaining the current number formed by the path from the root. At each leaf, add the current number to a running sum. Return the sum after the traversal.
Pro tip: Clarify that the tree may be empty and that the sum can exceed the integer range, so consider using a 64-bit integer or big integer if necessary. Also, mention that you can solve it iteratively with a stack to avoid recursion depth issues.
Confirm that each root-to-leaf path forms a number by concatenating digits, and we need the sum of all such numbers. Ask about edge cases like empty tree or single node.
Decide between recursive DFS or iterative stack-based DFS. Recursive is simpler but may hit recursion limit; iterative is more robust.
At each node, update the current number as current * 10 + node.val. Pass this down to children.
When a node has no children, add the current number to the total sum. Return the sum after traversal.
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.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.