1.8k views

The pre-order traversal of a binary search tree is given by $12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20$. Then the post-order traversal of this tree is

1. $2, 6, 7, 8, 9, 10, 12, 15, 16, 17, 19, 20$
2. $2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12$
3. $7, 2, 6, 8, 9, 10, 20, 17, 19, 15, 16, 12$
4. $7, 6, 2, 10, 9, 8, 15, 16, 17, 20, 19, 12$
edited | 1.8k views
0
+4
Clearly root is 12 ,and in post order root is last.So reject A.

Now if you know that left most element will be printed by postorder at first place and left most in inorder is smallest which is 2.

Directly we can give b as answer without construction:)
0

ans is B

so ans is B

edited by
Sorting the given sequence gives Inorder traversal.

Constructing tree gives Option B as the postorder traversal.
–1 vote
answer will be B . 100% sure
–1 vote
B is the answer of the above question.