[ Algorithm ] 96. Unique Binary Search Trees

Source From Here
Question
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.

How-To
First note that dp[k] represents the number of BST trees built from 1....k;

Then assume we have the number of the first 4 trees: dp[1] = 1 ,dp[2] =2 ,dp[3] = 5, dp[4] =14 , how do we get dp[5] based on these four numbers is the core problem here.

The essential process is: to build a tree, we need to pick a root node, then we need to know how many possible left sub trees and right sub trees can be held under that node, finally multiply them.

To build a tree contains {1,2,3,4,5}. First we pick 1 as root, for the left sub tree, there are none; for the right sub tree, we need count how many possible trees are there constructed from {2,3,4,5}, apparently it's the same number as {1,2,3,4}. So the total number of trees under "1" picked as root is dp[0] * dp[4] = 14. (assume dp[0] =1). Similarly, root 2 has dp[1]*dp[3] = 5 trees. root 3 has dp[2]*dp[2] = 4, root 4 has dp[3]*dp[1]= 5 and root 5 has dp[0]*dp[4] = 14. Finally sum the up and it's done.

Now, we may have a better understanding of the dp[k], which essentially represents the number of BST trees with k consecutive nodes. It is used as database when we need to know how many left sub trees are possible for k nodes when picking (k+1) as root.

view plaincopy to clipboardprint?

1. public int numTrees(int n) {  
2.     int [] dp = new int[n+1];  
3.     dp[0]= 1;  
4.     dp[1] = 1;  
5.     for(int level = 2; level <=n; level++)  
6.         for(int root = 1; root<=level; root++)  
7.             dp[level] += dp[level-root]*dp[root-1];  
8.     return dp[n];  
9. }