Unique Binary Search Trees

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本文探讨了如何计算给定整数n时，可以构造的不同结构的二叉搜索树的数量。通过递推公式实现了动态规划算法，有效地解决了这一经典问题。

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

Let C(n) be the number of distinct binary trees with n nodes. This is equal to the number of trees that have a root, a left subtree with j nodes, and a right subtree of (n-1)-j nodes, for each j. That is,

    C(n) = C(0)C(n-1) + C(1)C(n-2) + ... + C(n-1)C(0)

which is

http://cs.lmu.edu/~ray/notes/binarytrees/

// dp: bst[n] = bst[0]*bst[n-1] + bst[1]*bst[n-2] + ... + bst[n-1]*bst[0]

public class Solution {
    public int numTrees(int n) {
        int [] bst = new int[n+1];
        
        bst[0] = 1;     // empty subtree
        bst[1] = 1;
        for(int i=2; i<n+1; i++) {
            bst[i] = 0;
            for(int j=0; j<i; j++) {
                bst[i] += bst[j] * bst[i-j-1];
            }
        }
        
        return bst[n];
    }
}