Which of the following approaches correctly adapts the maximum path sum algorithm to handle this variant?
hard🎤 Interviewer Follow-up Q15 of Q15
Tree: Depth-First Search - Binary Tree Maximum Path Sum
Suppose the problem is modified so that the path can reuse nodes multiple times (i.e., cycles allowed), and node values can be negative. Which of the following approaches correctly adapts the maximum path sum algorithm to handle this variant?
AUse the same postorder traversal with global max tracking, ignoring negative gains as before.
BDetect cycles and use dynamic programming with memoization to avoid infinite loops, allowing repeated nodes in paths.
CConvert the tree to a graph and run Bellman-Ford algorithm to find the maximum sum path allowing cycles.
DUse depth-first search with pruning to explore all paths, but limit path length to avoid infinite loops.
Step-by-Step Solution
Solution:
Step 1: Understand the impact of allowing node reuse
Allowing cycles means paths can revisit nodes infinitely, so naive DFS or postorder traversal fails due to infinite loops.
Step 2: Identify correct approach
Dynamic programming with cycle detection and memoization prevents infinite recursion and correctly computes max sums with repeated nodes.
Final Answer:
Option B -> Option B
Quick Check:
Memoization with cycle detection handles repeated nodes safely [OK]
Quick Trick:Cycle detection and memoization needed for repeated nodes [OK]
Common Mistakes:
MISTAKES
Using same postorder traversal causes infinite loops
Bellman-Ford is for shortest paths, not max sums with negative cycles
DFS without cycle checks leads to infinite recursion
Trap Explanation:
PITFALL
Naive traversal ignores cycles, causing infinite loops; Bellman-Ford is not directly applicable for max path sums with negative values in trees.
Interviewer Note:
CONTEXT
Tests candidate's ability to adapt algorithms to graph variants and handle cycles.
Master "Binary Tree Maximum Path Sum" in Tree: Depth-First Search
3 interactive learning modes - each teaches the same concept differently
