[LeetCode]Unique Binary Search Trees

最新推荐文章于 2020-02-29 22:29:24 发布

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本文探讨了一种特定的组合数学问题，即计算不同形态的二叉搜索树的数量。通过建立递推方程和使用动态规划方法求解，文章详细介绍了如何利用Catalan数的性质来解决此类问题。

class Solution {
//a combinatorial mathematics problem 
//Firstly, this kind of problem generally can be modeled as famous number, here is catalan number
//Secondly, If we do not have such a background, I think it will be totally fine if we can find 
//the transform equation or the regular pattern.

//analysis:
//we pick a node i as a root in the BST, then the size of left subtree is i-1, and the size of 
//right tree is n-i, though the number in right tree may be different from 1 2 ... n-i, but actually
//it is equal to the sequence of that in such a problem.
//Now we should figure out that, if j is not equal to i, is there any duplicate(same) case?
//I think there will be no same case, because the root is different that will be enough to prove this.
//So, the equation will be f(size)=sum of f(i-1)*f(size-i), where 1<=i<=size. 
//initialize: f(0)=1, f(1)=1 
public:
	int DP(int n)
	{
		if(n <= 1)
			return 1;
		vector<int> f(n+1, 0);
		f[0] = f[1] = 1;
		for (int size = 2; size <= n; ++size)
		{
			for (int i = 1; i <= size; ++i)
			{
				f[size] += f[i-1]*f[size-i];
			}
		}
		return f[n];
	}
	int numTrees(int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		return DP(n);
	}
};

second time

class Solution {
public:
    int numTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > f(n, vector<int>(n, 0));
        for(int i = 0; i < n; ++i) f[i][i] = 1;

        for(int len = 2; len <= n; ++len)
        {
            for(int i = 0; i < n; ++i)
            {
                int j = len-1+i;
                if(j >= n) break;
                for(int k = i; k <= j; ++k)
                {
                    if(k == i) f[i][j] += f[k+1][j];
                    else if(k == j) f[i][j] += f[i][k-1];
                    else f[i][j] += (f[i][k-1]*f[k+1][j]);
                }
            }
        }
        return f[0][n-1];
    }
};