unique-binary-search-trees

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：

我们设置dp[n]是存储n个连续值的BST的数量

得到dp[n]:

设1为根节点，左子树有0个子节点，右子树有n - 1个子节点，bst个数为dp[0] * dp[n - 1](设dp[0] = 1);

设2为根节点，左子树有1个子树，右子树有n - 2个子树，bst个数为dp[1] * dp[n - 2]; 

一般情况下，我们将k设为根节点，所以左边的子树有k - 1个子节点，右边的子树有n - k个子节点，bst的个数是dp[k - 1] * dp[n - k];

dp[n] = d[0] * dp[n - 1] +…+ dp[k - 1] * dp[n - k] +…+ dp[n - 1] * dp[0];

class Solution {
public:
    int numTrees(int n) {
        vector<int> dp(n + 1, 0);
        dp[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int k = 1; k <= i; ++k) {
                dp[i] += dp[k - 1] * dp[i - k];
            }
        }
        return dp[n];
    }
};

 

转载于:https://www.cnblogs.com/lijingblog/p/11083796.html