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What is the time complexity of the optimal DFS with early stopping solution for the Path Sum problem on a binary tree with n nodes and height h?

medium🪤 Complexity Trap Q13 of Q15
Tree: Depth-First Search - Path Sum
What is the time complexity of the optimal DFS with early stopping solution for the Path Sum problem on a binary tree with n nodes and height h?
AO(h) because the recursion only goes down one path at a time
BO(n^2) because each node's sum is recalculated multiple times
CO(n) because in the worst case all nodes are visited once
DO(log n) assuming the tree is balanced
Step-by-Step Solution

  1. Step 1: Identify worst-case traversal

    In the worst case, the algorithm visits every node once to check all root-to-leaf paths.

  2. Step 2: Analyze recursion and early stopping

    Early stopping can prune some paths, but worst case still requires full traversal, so time is O(n).

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    Each node visited once -> O(n) time [OK]
Quick Trick: Worst case requires visiting all nodes once [OK]
Common Mistakes:
MISTAKES
  • Confusing height with total nodes
  • Assuming early stopping always reduces complexity
Trap Explanation:
PITFALL
  • Option A looks plausible because recursion depth is h, but all nodes may be visited, so O(n) is correct.
Interviewer Note:
CONTEXT
  • Checks understanding of traversal complexity vs recursion depth.
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