Tree: Depth-First Search - Binary Tree Preorder TraversalWhat is the time complexity of the Morris preorder traversal algorithm on a binary tree with n nodes?AO(n log n) due to repeated threadingBO(n^2) because of nested loops in threadingCO(n) because each edge is visited at most twiceDO(log n) since traversal is balancedCheck Answer
Step-by-Step SolutionSolution:Step 1: Analyze edge visitsEach edge is visited at most twice: once to create thread, once to remove it.Step 2: Total operations proportional to nodesSince edges are n-1, total operations are O(n).Final Answer:Option C -> Option CQuick Check:Linear time traversal with threading [OK]Quick Trick: Each edge visited twice -> O(n) [OK]Common Mistakes:MISTAKESAssuming nested loops cause O(n^2) complexityTrap Explanation:PITFALLCandidates confuse inner loops with repeated full traversals, inflating complexity.Interviewer Note:CONTEXTTests understanding of Morris traversal time complexity.
Master "Binary Tree Preorder Traversal" in Tree: Depth-First Search3 interactive learning modes - each teaches the same concept differentlyTry ItSolutionTrace
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