leetcode 96. Unique Binary Search Trees

1.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

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2.思路

DP思想，设Tree[n]表示序列{1，2.。。n}构成的二叉树数目，那么显然问题可以分解为将每个数当作root，求所有情况的和。当把i当作root时，二叉树数目为Tree[i-1]*Tree[n-i].就是左子数数目和又子树数目相乘。那么状态转移方程为：Tree[n]=Tree[0]*Tree[n-1]+Tree[1]*Tree[n-2]+....+Tree[n-1]*Tree[0] ;初始条件为：Tree[0]=Tree[1]=1;

3.代码

int numTrees(int n) {    int* Tree=new int[n+1]();//调用默认构造函数初始化为0    Tree[0] = Tree[1] = 1;        for(int i=2; i<=n; ++i) {    for(int j=1; j<=i; ++j) {    Tree[i] += Tree[j-1] * Tree[i-j];    }    }    return Tree[n];    }