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Let $T(n)$ be the number of different binary search trees on $n$ distinct elements.

Then $T(n) = \sum_{k=1}^{n} T(k-1)T(x)$, where $x$ is 

  1. $n-k+1$
  2. $n-k$
  3. $n-k-1$
  4. $n-k-2$
asked in DS by Veteran (59.4k points)
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3 Answers

+34 votes
Best answer

The summation is for each node, if that node happens to be the root. When a node is root, it will have $(k-1)$ nodes on the left sub tree ($k$ being any number) and correspondingly $(n-k)$ elements on the right sub tree.  So, we can write recurrence $T(k-1) * T(n-k) $ for the number of distinct binary search trees, as the numbers on left and right sub trees form BSTs independent of each other and only a difference in one of the sub trees produces a difference in the tree. Hence, answer is B

Knowing the direct formula can also help in getting the answer but is not recommended. 

http://gatecse.in/wiki/Number_of_Binary_trees_possible_with_n_nodes

answered by Veteran (342k points)
edited by
0

So, we can write recurrence T(k-1) * T(n-k) for the number of distinct binary search trees, as the numbers on left and right sub trees form BSTs independent of each other.

Please explain this preferably with an example.

@Arjun Sir

+2

watch this video for clarity. 

+38 votes
left subtree + root + right subtree = total node
(k-1) + 1 + x = n
x = n - k
answered by Veteran (54.8k points)
+29 votes

Option B

answered by Boss (25.7k points)
edited by
0
Thank U very Much !!! I got this Knot !!!
0
thank u sir .....
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