How do you recover a BST where exactly two nodes were swapped?
PICTURE THIS: ARRAY IN MEMORY
Index starts at 0. Scan once for max — O(n).
Simple meaning
Inorder should be sorted
WHY — BST instead of guessing?
Why interviewers care about BST:
question about BST.
trade-offs, and what you would actually do on a DSA project - not buzzwords.
Name the idea, why it exists, then one short example.
End with when you use it and one common pitfall.
STEPS — What happens step by step?
Before you speak the answer, walk the interviewer through these steps:
- 1Define it
Inorder should be sorted
- 2find the one or
two inversions (first dip and last dip) and swap those node values.
- 3Time is O(n) and
space is O(h) with recursion, or O(1) extra with Morris traversal.
- 4Recreating the tree from
sorted keys would use extra arrays.
- 5Common mistake
What juniors usually get wrong.
- 6Close
When you pick this over the alternative.
EXAMPLE — See it in action
Here's a short line you can speak, broken into clear beats:
Note: Adapt this scaffold to your own project — keep it under 60–90 seconds.
Key takeaway
Inorder should be sorted find the one or two inversions (first dip and last dip) and swap those node values.