[leetcode]96. Unique Binary Search Trees(Java)
阿新 • • 發佈:2019-02-08
https://leetcode.com/problems/unique-binary-search-trees/#/description
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
package go.jacob.day712; import org.junit.Test; /** * 96. Unique Binary Search Trees * * @author Jacob * */ public class Demo1 { /* * 只要確定根節點,那麼左右子樹節點個數就能確定 * * 解析: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 int numTrees(int n) { if (n <= 0) return 0; int[] res = new int[n + 1]; res[0] = 1; res[1] = 1; for (int i = 2; i <= n; i++) { for (int j = 1; j <= i; j++) { res[i] += res[j - 1] * res[i - j]; } } return res[n]; } }