I'm using the definition of the height of a tree as the longest possible path from the root to a leaf by its number of edges, e.g. a tree of 2 nodes has a height of 1. With that in mind, what would be the number of binary search trees (don't have to be balanced) that have the maximum possible height for n-nodes?
So far, I know that the maximum height of any binary search tree of n-nodes is n-1 since we're counting edges. So for example, a binary search tree of 3 nodes has a maximum height of h = 3-1 = 2 and there are 4 possible trees with n-nodes that have the maximum height of 2.
I also know that for n-nodes, there are (2n)!/((n+1)!n!) possible trees.
I'm able to find this manually for trees where n is small but would like to find a formula for any n.
EDIT:
The BST definition I am working with defines a Binary Search tree as a binary tree where each node has at most 2 children and does NOT need to be nearly-complete, so you could have a tree of 3 nodes with a height of 2. It also satisfies the BST property where all nodes in the left subtree of a node x has keys smaller than x's key and all nodes in the right subtree of a node x has keys larger than x's key.
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1$\begingroup$ Have you plugged in the values you've found for small $n$ in the Online Encyclopedia of Integer Sequences? $\endgroup$Yuval Filmus– Yuval Filmus2018年02月17日 09:32:37 +00:00Commented Feb 17, 2018 at 9:32
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$\begingroup$ In a binary search tree, every internal node has two children. In particular there is no binary search tree with only two nodes. A binary search tree with three nodes has height 1, and in general the maximum height of a binary search tree with $n$ nodes is $n-2$. There are 2ドル^{n-3}$ binary trees of this height, since at every internal node but the last one we choose which side is a leaf. $\endgroup$Yuval Filmus– Yuval Filmus2018年02月17日 14:17:53 +00:00Commented Feb 17, 2018 at 14:17
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$\begingroup$ @YuvalFilmus The definition I'm working with considers a binary search tree to be a binary tree with at most 2 children, but it does not need to be nearly-complete. It also has the BST property. So I can have a 3-node tree of height 2. $\endgroup$Paradox– Paradox2018年02月19日 21:35:01 +00:00Commented Feb 19, 2018 at 21:35
1 Answer 1
The number of trees with $n$ nodes of height $n-1$ is 2ドル^{n-1}$. Indeed, every internal node has exactly one child, which can either be the left child or the right child. Since there are $n-1$ internal nodes, this gives 2ドル^{n-1}$ options.
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