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96. Unique Binary Search Trees(dp) #162

Open Shawngbk opened 7 years ago

Shawngbk commented 7 years ago

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.

public class Solution { public int numTrees(int n) { int [] dp = new int[n+1]; dp[0] = dp[1] = 1; for(int level = 2; level <= n; level++) { for(int root = 1; root <= level; root++) { dp[level] += dp[root-1] * dp[level-root]; } } return dp[n]; } }

Shawngbk commented 7 years ago

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