96. Unique Binary Search Trees 类别:动态规划 难度:medium
2016-11-06 22:49
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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.
思路:
利用动态规划算法,并且根据二叉搜索树的特点,用dp[1...n][1...n]来记录结果,dp[m]
表示将m到n组成二叉搜索树的形态个数,在m和n之间取k为根的话,则m到k-1组成左子树,而k+1到n组成右子树,于是可以写出转移方程dp[m]
= dp[m][k - 1] * dp[k + 1]
。
程序:
class Solution {
public:
int numTrees(int n) {
vector<vector<int> > dp(n + 1,vector<int>(n + 1,1));
for(int i = 1;i <= n - 1;i++)
{
for(int j = 1;j + i <= n;j++)
{
dp[j][j + i] = 0;
for(int k = j + 1;k <= j + i - 1;k++)
dp[j][j + i] += dp[j][k - 1] * dp[k + 1][j + i];
dp[j][j + i] += (dp[j + 1][j + i] + dp[j][j + i - 1]);
}
}
return dp[1]
;
}
};
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:
利用动态规划算法,并且根据二叉搜索树的特点,用dp[1...n][1...n]来记录结果,dp[m]
表示将m到n组成二叉搜索树的形态个数,在m和n之间取k为根的话,则m到k-1组成左子树,而k+1到n组成右子树,于是可以写出转移方程dp[m]
= dp[m][k - 1] * dp[k + 1]
。
程序:
class Solution {
public:
int numTrees(int n) {
vector<vector<int> > dp(n + 1,vector<int>(n + 1,1));
for(int i = 1;i <= n - 1;i++)
{
for(int j = 1;j + i <= n;j++)
{
dp[j][j + i] = 0;
for(int k = j + 1;k <= j + i - 1;k++)
dp[j][j + i] += dp[j][k - 1] * dp[k + 1][j + i];
dp[j][j + i] += (dp[j + 1][j + i] + dp[j][j + i - 1]);
}
}
return dp[1]
;
}
};
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