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Unique Binary Search Trees

时间:2015-10-20 19:28:31      阅读:157      评论:0      收藏:0      [点我收藏+]

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.

看得别人的代码,然后写出来的 :http://www.2cto.com/kf/201312/262420.html

public class UniqueBinarySearchTrees {

    public static void main(String[] args) {
        int n = (int)(Math.random()*100); // 随机生成数字
        System.out.println(n);
        int num = numTrees(n);
        System.out.println(num);
    }

    private static int numTrees(int n) {
        if (n == 0 || n == 1) {
            return n;
        }
        else {
            // 以i为根节点时,其左子树构成为[0,...,i-1],其右子树构成为[i+1,...,n]构成
            // 因此,dp[i] = sigma(dp[0...k] * dp[k+1...i]) 0 <= k < i - 1 
            // dp[3] = dp[0] * dp[2] + dp[1] * dp[1] + dp[2] * dp[0] 
            int[] num = new int[n+1];
            num[0] = num[1] = 1;
            for (int i = 2; i < num.length; i++) {
                num[i] = 0;
                for (int j = 0; j < i; j++) {
                    num[i] = num[i] + num[j] *  num[i-j-1]; 
                }
            }
            return num[n];
        }
    }

}

 

Unique Binary Search Trees

原文:http://www.cnblogs.com/tf-Y/p/4895693.html

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