高度为h可以形成多少种不同的二叉树？

Question

How many types of distinct Binary Tree can be formed with a height of h? if we only know the height of binary tree, and we regard root-left and root-right as the same tree structure, this means if the height of binary tree equal to 1, we can formed 2 different trees structure: root - leftchild/rightchild; root - leftchild - rightchild

Answer 1

Let's there are S[h] ways to build a tree with a height of h ( (h)-tree ).
We can compose (h-tree) from root node and from:
- left (h-1)-tree and any right tree with height 0..h-2
- two (h-1)-trees
- right (h-1)-tree and any left tree with height 0..h-2

So recursive formula is

 S[h] = S[h-1] * Sum{k=0..h-2} (S[k]) + 
        S[h-1] * S[h-1] + 
        S[h-1] * Sum{k=0..h-2} (S[k]) = 
        S[h-1] * (S[h-1]  + 2 * Sum{k=0..h-2} (S[k]))

Now with S[0] = 1, S[1] = 1 we can find all values with recursion.

Example: S[3] = S[2] * (S[2] + 2*(S[0] + S[1])) = 3 * (3 + 2 * (1 + 1)) = 21

Note that the same values will be computed again and again, so it is worth to use dynamic programming:
- or top-down memoization - when we write once calculated value S[k] to the table and use it later, it is very simple modification of recursion
- or bottom-up approach - here we can use additional table with cumulative sums.

To check: S[5] = 457653