Unique Binary Search Trees
2013-07-17 22:17
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考虑根节点i可组成的BST的数量,
小于i的(i-1)个数可以组成N颗BST,大于i的(n-i)个数可以组成M颗BST,
因此以i为根节点的BST共有M*N颗。
小于i的(i-1)个数可以组成N颗BST,大于i的(n-i)个数可以组成M颗BST,
因此以i为根节点的BST共有M*N颗。
class Solution { public: int numTrees(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<int> f(n+2,0); f[0]=1; f[1]=1; for(int i=2;i<=n;++i){ for(int j=1;j<=i;++j){ f[i]+=f[j-1]*f[i-j]; } } return f ; } };
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