Unique Binary Search Trees

[江湖小虾米东方未明]

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



class Solution {
public:
int numTrees(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> v(n+1, 0);
v[0] = 1;
v[1] = 1;
v[2] = 2;
if(n<=2)
return v[n];

for(int i=3; i<=n; i++)
for(int j=1; j<=i; j++)
v[i] += v[j-1]*v[i-j];
return v[n];
}
};
求人解释下这个Dp方程式的意义啊。。看不懂

[Hony杨]

v[n] is total of n unique BST's 1 2 ... i ... n, we regard i as the root of tree, 1 ..2..(i-1) is left sub tree, (i+1) ..n is right sub tree. so, v[n]=sum{v[i-1]*v[n-i]}, i=[1,n], v[0]=v[1]=1, ex: Given n =3, v[3]=sum{v[0]*v[2],v[1]*v[1],v[2]*v[0]} = 5 v[2]=sum{v[0]*v[1],v[1]*v[0]} =2 当我们取出一个数x作为根节点后，那么小于x的数必然会成为左子树，大于x的数必然会成为右子树，也就是说对于BST，只和数的个数有关

[江湖小虾米东方未明]

引用 1 楼 HuiyuYang_fish 的回复:

v[n] is total of n unique BST's 1 2 ... i ... n, we regard i as the root of tree, 1 ..2..(i-1) is left sub tree, (i+1) ..n is right sub tree. so, v[n]=sum{v[i-1]*v[n-i]}, i=[1,n], v[0]=v[1]=1, ex: Given n =3, v[3]=sum{v[0]*v[2],v[1]*v[1],v[2]*v[0]} = 5 v[2]=sum{v[0]*v[1],v[1]*v[0]} =2 当我们取出一个数x作为根节点后，那么小于x的数必然会成为左子树，大于x的数必然会成为右子树，也就是说对于BST，只和数的个数有关

谢谢MTK_娃娃鱼的帮助，看了你的解释后豁然开朗.