LeetCode 题解（27）: Unique Binary Search Trees

最新推荐文章于 2025-08-21 20:43:22 发布

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最新推荐文章于 2025-08-21 20:43:22 发布

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分类专栏：算法文章标签： algorithm leetcode Unique Binary Search

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本文详细介绍了如何通过递归方法计算给定范围内不同长度的二叉搜索树的数量，并提供了多种编程语言实现代码。

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题目：

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

题解：

将树分成两半，[1,i-1] - i - [i+1,n]，i属于[1,n]，且为根节点，则左边都比它小，右边都比它大。以i为根节点的子树个数 = 左子树个数 * 右子树个数。递归下去。若start >= end 返回1.

class Solution {
public:
    int numTrees(int n) {
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            int num = numTreesRecursion(1, i-1) * numTreesRecursion(i+1, n);
            sum += num;
        }
        return sum;
    }
    
    int numTreesRecursion(int start, int end)
    {
        if(start >= end)
            return 1;

        int sum = 0;
        for(int i = start; i <= end; i++)
        {
            int num = numTreesRecursion(start, i-1) * numTreesRecursion(i+1, end);
            sum += num;
        }
        return sum;
    }
};

八个月后第二次做：

class Solution {
public:
    int numTrees(int n) {
        if(!n)
            return 0;
        
        int number = 0;
        
        for(int i = 1; i <= n; i++) {
            number += numBST(1, i-1) * numBST(i+1, n);
        }
        
        return number;
    }
    
    int numBST(int start, int end) {
        if(start >= end)
            return 1;
        int sum = 0;
        for(int j = start; j <= end; j++) {
            sum += numBST(start, j-1) * numBST(j+1, end);
        }
        return sum;
    }
};

Java版：

public class Solution {
    public int numTrees(int n) {
        if(n == 0)
            return 0;
        int number = 0;
        for(int i = 1; i <= n; i++) {
            number += numBST(1, i-1) * numBST(i+1, n);
        }
        return number;
    }
    
    public int numBST(int start, int end) {
        if(start >= end)
            return 1;
        int num = 0;
        for(int j = start; j <= end; j++) {
            num += numBST(start, j-1) * numBST(j+1, end);
        }
        return num;
    }
}

Python版：

class Solution:
    # @return an integer
    def numTrees(self, n):
        if n == 0:
            return 0
        number = 0
        for i in range(1,n+1):
            number += self.numBST(1,i-1) * self.numBST(i+1,n)
        return number
        
    def numBST(self, start, end):
        if start >= end:
            return 1
        num = 0
        for j in range(start, end+1):
            num += self.numBST(start, j-1) * self.numBST(j+1, end)
        return num