Leetcode: Unique Binary Search Trees

最新推荐文章于 2025-08-08 00:00:33 发布

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最新推荐文章于 2025-08-08 00:00:33 发布

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分类专栏： Leetcode 文章标签： leetcode

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本文介绍了一种计算给定整数n时，有多少种结构上独一无二的二叉搜索树的方法。通过递推公式C(n)=C(0)C(n-1)+C(1)C(n-2)+...+C(n-1)C(0)，给出了一种高效的计算方式，并提供了完整的Java实现代码。

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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

The first few terms:

    C(0) = 1
    C(1) = C(0)C(0) = 1
    C(2) = C(0)C(1) + C(1)C(0) = 2
    C(3) = C(0)C(2) + C(1)C(1) + C(2)C(0) = 5
    C(4) = C(0)C(3) + C(1)C(2) + C(2)C(1) + C(3)C(0) = 14

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