[leetcode-96-Unique Binary Search Trees]

最新推荐文章于 2019-06-07 22:20:45 发布

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最新推荐文章于 2019-06-07 22:20:45 发布

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原文链接：http://www.cnblogs.com/hellowooorld/p/7041062.html

本文介绍了一种使用动态规划的方法来计算给定整数n时，可以构造的不同形态的二叉搜索树的数量。通过将问题分解为子问题，并利用递推公式F(n)=F(0)*F(n-1)+...+F(n-1)*F(0)，有效地解决了问题。

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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：

/**
 * Taking 1~n as root respectively:
 *      1 as root: # of trees = F(0) * F(n-1)  // F(0) == 1
 *      2 as root: # of trees = F(1) * F(n-2) 
 *      3 as root: # of trees = F(2) * F(n-3)
 *      ...
 *      n-1 as root: # of trees = F(n-2) * F(1)
 *      n as root:   # of trees = F(n-1) * F(0)
 *
 * So, the formulation is:
 *      F(n) = F(0) * F(n-1) + F(1) * F(n-2) + F(2) * F(n-3) + ... + F(n-2) * F(1) + F(n-1) * F(0)
 */

int numTrees(int n) 
{
  vector<int>dp(n+1,0);
  dp[0] = dp[1] = 1;
  for(int i=2;i<=n;i++)
  {
    dp[i] =0;
    for(int j = 1;j<=i;j++)
    {
      dp[i]+= dp[j-1] * dp[i-j];
    }
  }
  return dp[n];
}