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It works correctly: if v1 is an ancestor of v2, then the algorithm will traverse the tree until root is the node containing v1. At this point root.data < v1 is not true and root.data > v1 is not true, therefore both of the if-statements are bypassed and root is returned, which is the correct behavior.
No, if v1 is an ancestor of v2, v1 will be the lca. (this confused me for quite some time and I hope they change the challenge description to account for this.)
Binary Search Tree : Lowest Common Ancestor
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It works correctly: if v1 is an ancestor of v2, then the algorithm will traverse the tree until root is the node containing v1. At this point root.data < v1 is not true and root.data > v1 is not true, therefore both of the if-statements are bypassed and root is returned, which is the correct behavior.
can u help me to bug what is wrong with testcase 2 for my code.
If v1 is the ancestor of v2, then should the LCA = parent of v1 instead of v1?
No, if v1 is an ancestor of v2, v1 will be the lca. (this confused me for quite some time and I hope they change the challenge description to account for this.)
Agreed, it confused me too until I read your comment! Thanks :)
How can v1 be the ancestor of itself? I am still confused.
Or, are you referring to the correct answer that the question will accept?
Thanks!