96. Unique Binary Search Trees

最新推荐文章于 2021-05-17 12:15:23 发布

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本文介绍了一种生成所有结构上唯一的二叉搜索树（BST）的算法，这些BST存储从1到n的值。通过递归地选择根节点并确定左右子树的元素数量来解决该问题。给出的代码实现了这一算法，并展示了当n为3时，可以生成5棵不同的二叉搜索树。

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,
Given n = 3, your program should return all 5 unique BST's shown below.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

The idea is Suppose we know the solution from 1 to n-1, and we want to know about n;

F(n) = 0;

We could choose a number X from 1 to n to be the root, and all the numbers in the left subtree i < X and in the right tree i > X

There are (i - 1) number in the left tree and (n - i) number in the right tree.

F(n) += F(i - 1) * F(n - i) for i from 1 to N

Code:

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

in the above code k stands for how many number in the left tree and (i - k - 1) stands for how many number in the right tree.