Given n, how many structurally unique BST’s (binary search trees) that store values 1…n?

For example, Given n = 3, there are a total of 5 unique BST’s.

1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3

**Analysis**

Let count[i] be the number of unique binary search trees for i. The number of trees are determined by the number of subtrees which have different root node. For example,

i=0, count[0]=1 //empty tree i=1, count[1]=1 //one tree i=2, count[2]=count[0]*count[1] // 0 is root + count[1]*count[0] // 1 is root i=3, count[3]=count[0]*count[2] // 1 is root + count[1]*count[1] // 2 is root + count[2]*count[0] // 3 is root i=4, count[4]=count[0]*count[3] // 1 is root + count[1]*count[2] // 2 is root + count[2]*count[1] // 3 is root + count[3]*count[0] // 4 is root .. .. .. i=n, count[n] = sum(count[0..k]*count[k+1...n]) 0 <= k < n-1

Use dynamic programming to solve the problem.

**Java Solution**

public int numTrees(int n) { int[] count = new int[n + 1]; count[0] = 1; count[1] = 1; for (int i = 2; i <= n; i++) { for (int j = 0; j <= i - 1; j++) { count[i] = count[i] + count[j] * count[i - j - 1]; } } return count[n]; } |

Check out how to get all unique binary search trees.

def fun():

(2n)!/(n+1)!n!

Catalan number of ways

Is there a typo here?

i=2, count[2]=count[0]*count[1] // 0 is root

+ count[1]*count[0] // 1 is root

There is no node with val 0, so 0 cannot be root. Comments should be // 1 is root

// 2 is root.