Catlan nos difficulty

#1

while calculating no of unlabelled binary trees why they are same as no of bsts as in case of unlabelled bt there is no condition to have i-1 node on left side and n-i nodes on right side

#2

@Namanjain123
In case of BST for given n nodes, you can see that all the arrangements are different structure-wise, i.e. the structure of all the possibilites will be unique. (take n=3 and draw all the possibilities, and then just draw them again without giving the value to the node).

Similarly, in case of unlabelled binary trees, only the arrangement or structure matters.
So the answer will be same for both.

#3

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