Which modification to the BFS-based maximum depth algorithm correctly computes the maximum depth in this scenario?
hard🎤 Interviewer Follow-up Q15 of Q15
Tree: Depth-First Search - Maximum Depth of Binary Tree
Suppose the binary tree nodes can have an arbitrary number of children (not just two). Which modification to the BFS-based maximum depth algorithm correctly computes the maximum depth in this scenario?
AReplace 'if node.left' and 'if node.right' checks with iterating over 'node.children' list and enqueue each child.
BKeep the same code but only enqueue 'node.left' and 'node.right' if they exist, ignoring other children.
CUse DFS recursion only, as BFS cannot handle nodes with more than two children.
DModify the code to sum depths of all children instead of taking the maximum.
Step-by-Step Solution
Solution:
Step 1: Understand the problem extension
Nodes now have arbitrary children, so left/right pointers are insufficient.
Step 2: Modify BFS to handle multiple children
Iterate over 'node.children' list and enqueue each child to explore all branches correctly.
Step 3: Evaluate other options
Ignoring children misses nodes; DFS can work but BFS is still valid; summing depths is incorrect for max depth.
Final Answer:
Option A -> Option A
Quick Check:
Enqueue all children ensures full traversal for max depth [OK]
Quick Trick:Enqueue all children nodes to handle arbitrary branching [OK]
Common Mistakes:
MISTAKES
Ignoring extra children
Assuming binary tree structure
Summing depths instead of max
Trap Explanation:
PITFALL
Ignoring children beyond left/right seems simpler but breaks correctness.
Interviewer Note:
CONTEXT
Tests adaptability of BFS approach to generalized tree structures.
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