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What is the space complexity of the optimal recursive approach to invert a binary tree, assuming the tree has n nodes and height h?

medium🪤 Complexity Trap Q13 of Q15
Tree: Depth-First Search - Invert Binary Tree
What is the space complexity of the optimal recursive approach to invert a binary tree, assuming the tree has n nodes and height h?
AO(n) because each node's children pointers are swapped individually
BO(n) due to storing all nodes in a queue during traversal
CO(1) because inversion is done in-place without extra data structures
DO(h) due to recursion stack depth proportional to tree height
Step-by-Step Solution

  1. Step 1: Identify auxiliary space usage in recursion

    The algorithm uses recursion, so the call stack depth is proportional to the height h of the tree.

  2. Step 2: Distinguish between in-place operations and recursion stack

    Swapping pointers is in-place (O(1) per node), but recursion stack space is O(h), not O(n).

  3. Final Answer:

    Option D -> Option D

  4. Quick Check:

    Recursion stack dominates space, proportional to tree height h [OK]
Quick Trick: Recursion stack space depends on tree height, not node count [OK]
Common Mistakes:
MISTAKES
  • Confusing in-place operation with O(1) total space
  • Assuming queue-based BFS space applies to recursive DFS
  • Counting node swaps as extra space
Trap Explanation:
PITFALL
  • Many think in-place means O(1) total space ignoring recursion stack, or confuse BFS queue space with DFS recursion space.
Interviewer Note:
CONTEXT
  • Tests understanding of recursion stack space vs in-place pointer swaps.
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