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Adobe Interview Question


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of 0 vote

I believe you mean two binary "search" trees?

- Anonymous March 11, 2010 | Flag Reply
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of 0 votes

If it's two binary search trees, then a possible solution is to do Inorder traversal. If not search tree, any traversal will do. After such a traversal, the data from the binary tree can be stored and then the two arrays can be compared for similarity in data.

- Anonymous March 11, 2010 | Flag
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0
of 0 vote

Not necessarily BST,
int sameTrees(Node* a, Node*b)
{
if (a == null && b == null) return (true);
else if (a !=null && b != null)
return ((a->data == b->data)&& (sameTree(a->left,b->left) && (sameTree(a->right, b->right)));
else
return false;

- tetura March 12, 2010 | Flag Reply
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0
of 0 votes

plzzzzzz comment on this soln

- sunny March 13, 2010 | Flag
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0
of 0 votes

This is a correct solution

- Dot Divx April 16, 2010 | Flag
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0
of 0 votes

This is *not* correct. You cannot assume that the trees are structured exactly the same way. The keys for two trees could be identical but the way they are structured may be different.
Do inorder traversal of the first tree and hash the keys. Now do inorder traversal of the second tree and make sure that all the keys encountered belong in the hash table.

- Anonymous May 19, 2010 | Flag
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0
of 0 votes

thats the correct solution. with your approach you are just checking the contents. consider example
1. root : b
left node : a
Right node : c
2 root : a
right node : b
right node of b : c

- Anonymous August 06, 2010 | Flag
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0
of 0 vote

thats the correct solution. with your approach you are just checking the contents. consider example
a a
/ \ AND \
b c b
\
c

- Anonymous August 06, 2010 | Flag Reply
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0
of 0 vote

l

- mail2ankitmalhotra October 26, 2010 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

l

- mail2ankitmalhotra October 26, 2010 | Flag Reply


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