【leetcode】 Unique Binary Search Trees (middle)☆

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

 

 

思路：因为是奔着动态规划强化来做的，所以方法肯定是动态规划。而且题目里连个树的定义都没给，肯定不用真的把树建出来。

首先说明下面我的思路是个很烂的思路，但是AC了

设f[a][b]表示若以c为头结点，有a个小于c的数和b个大于c的数时 可以表示的方式数，则

f[a][b + 1] = f[a][0]*(f[0][b] + f[1][b-1] +...+f[b-1][1] + f[b][0])

f[0][0] = 1

f[0][1] = 1

f[a][b] =f[b][a]

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

class Solution {
public:
    int numTrees(int n) {
        if(n == 0 || n == 1)
        {
            return n;
        }
        //f[a][b]表示若以c为头结点，有a个小于c的数和b个大于c的数时 可以表示的方式数
        vector<vector<int> > f(n, vector<int>(n, 0));
        f[0][0] = 1;
        f[0][1] = 1;
        f[1][0] = 1;

        for(int total = 3; total <= n; total++)
        {
            int less = (total - 1) / 2;
            for(int i = less; i >= 0; i--)
            {
                int ways = 0;
                int large = total - 1 - i;
                for(int j = 0; j <= large - 1; j++)
                {
                    ways += f[j][large - j - 1];
                }
                ways *= f[i][0];
                f[i][large] = ways;
                f[large][i] = ways;
            }
        }

        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            ans += f[i][n - 1 - i];
        }
        return ans;
    }
};

int main()
{
    Solution s;
    int ans = s.numTrees(3);

    return 0;

}

 

看别人的思路，发现自己傻×了。

idea: choose i to be the root, then there i -1 numbers in the left child tree, n - i numbers in the right child tree, calculate child tree recursively

f(0) = 1

f(1) = 1

f(2) = 2

f(n) = f(n-1) * f(0) + f(n-2) * f(1) + f(n-3) * f(2) + ... + f(0) * f(n-1)

int numTrees(int n) 
{
    if(n<0) return 0;

    vector<int> result;
    result.push_back(1);
    result.push_back(1);

    for(int k=2;k<=n;k++)
    {
        int sum=0;
        for(int i=1;i<=k;i++)
        {
            sum+=(result[i-1]*result[k-i]);
        }
        result.push_back(sum);
    }
    return result[n];
}

 

还有用数学方法的： Catalan number  C(i+1) = C(i) * 2 * ( 2 * i - 1 ) / (i + 1);

令h(0)=1,h(1)=1，catalan数满足递推式：

h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)

class Solution {
public:
    int numTrees(int n) {
        long long ans = 1;
        for(int i = 1; i <= n; ++ i)
            ans  = ans * 2 * (2 * i - 1) / (i + 1);
        return (int) ans;
    }
};