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What is the time complexity of the Morris inorder traversal algorithm on a binary tree with n nodes?

medium🪤 Complexity Trap Q13 of Q15
Tree: Depth-First Search - Binary Tree Inorder Traversal
What is the time complexity of the Morris inorder traversal algorithm on a binary tree with n nodes?
AO(n) because each edge is visited at most twice during threading and unthreading.
BO(n \times h) where h is tree height due to repeated traversal of left subtrees.
CO(n \log n) due to repeated searching for predecessors.
DO(n^2) in worst case when tree is skewed and predecessor search is costly.
Step-by-Step Solution

  1. Step 1: Analyze predecessor visits

    Each node's left subtree is traversed to find the predecessor, which can take up to h steps where h is the height of the tree.

  2. Step 2: Count total operations

    In skewed trees, this leads to O(n x h) time complexity due to repeated traversal of left subtrees.

  3. Final Answer:

    Option B -> Option B

  4. Quick Check:

    Worst-case time complexity is O(n x h) [OK]
Quick Trick: Predecessor search can take O(h) per node in skewed trees [OK]
Common Mistakes:
MISTAKES
  • Assuming predecessor search is O(1) per node
  • Confusing threading effect with total complexity
  • Ignoring skewed tree worst case
Trap Explanation:
PITFALL
  • Candidates often think Morris traversal is always O(n), but worst case can be O(n x h) due to predecessor searches.
Interviewer Note:
CONTEXT
  • Tests understanding of Morris traversal's worst-case time complexity.
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