Question

Consider I have an array [3,18,15,25,26], how many possible binary search trees can be formed from it?

Answer 1

After looking at the question linked by MicSim, I still wasn’t satisfied with that, so I began looking at it myself. Here’s what I came up with…

Each tree can be thought of as two trees with a parent root node. If you know the number of possible combinations of the two children branches separately, the total combinations having that root node is the product of the children combinations.

We can begin building up a higher count solution by solving the lower count instances first.

I will use C(n) to represent the total possible combinations of n nodes, the Catalan Number.

Hopefully these two are obvious:

C(0) = 1
C(1) = 1

C(2) is also fairly obvious, but it can be built, so let’s do that. There are two ways to choose the root node. One leaves a child count (left:right) of 1:0 and the other 0:1. So, the first possibility is C(1)*C(0) = 1*1 = 1. And the second is C(0)*C(1) = 1*1 = 1. Summing those together gives us

C(2) = 2

Nothing too exciting yet. Now let’s do 3 nodes. There are 3 ways to choose the root node and, hence, 3 child groupings. Your possible groups are 2:0, 1:1 and 0:2.

Based on our prior defintions, C(3) can be written as C(2)*C(0) + C(1)*C(1) + C(0)*C(2) = 2*1 + 1*1 + 1*2 = 2+1+2 = 5.

C(3) = 5

4 nodes has child groupings of 3:0, 2:1, 1:2 and 0:3. So, C(4) can be written as C(3)*C(0) + C(2)*C(1) + C(1)*C(2) + C(0)*C(3) = 5*1 + 2*1 + 1*2 + 1*5 = 5+2+2+5 = 14.

C(4) = 14

Hopefully, two things are beginning to become apparent. First, this will start becoming cumbersome fairly soon. Second, that what I have described, in a fairly long-winded fashion, is the recurrence relation representation on the wiki page.

I don’t know if this helps, but it helped me to go through the exercise, so I thought I’d share. I wasn’t trying to recreate the recurrence relation when I started, so it was good that my results matched one of the existing methods.