Leetcode: 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

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1. f(n) = \sum_{i=1}^n f(i-1) * f(n-i), f(0) = 1, f(1) = 1;
2. This is nature for DP (see below).
3. f(n) = (2n)!/n!*(n+1)!, called Catalan number

   class Solution{
   public:
           int numTrees(int n)
           {
               vector<int> num(n+1, 0);
               num[0] = 1;
               if (n > 0)
                   num[1] = 1;
               for(int i = 2; i < n+1; i++)
               for(int j = 0; j < i; j++)
               {
                   num[i] += num[j]*num[i-j-1];
               }
               return num[n];
           }
           
   };

   解法二：递归来做，不过有很多重复计算

   //from stanford
   
   public class Solution { public int numTrees(int numKeys) {
   
       if(numKeys <= 1)
           return 1;
       // there will be one value at the root, with whatever remains 
      // on the left and right each forming their own subtrees. 
     // Iterate through all the values that could be the root... 
       int sum = 0;
       int left, right, root;
   
       for(root = 1; root <= numKeys; root++) {
           left = countTrees(root - 1);
           right = countTrees(numKeys - root);
   
           sum += left * right;
       }
   
       return sum;
   }
   }

   Use DP to solve this question. For any node i between node 1 to node n, f(n) += f(i-1)*f(n-i).
   
   int numTrees(int n) {
       int num, *f = (int*)malloc((n+1)*sizeof(int));
       memset(f, 0, (n+1)*sizeof(int));
   
       f[0] = 1;
       f[1] = 1;
   
       for (int j = 2; j <= n; j++) {
           for (int i = 1; i <= j; i++) {
               f[j] += f[i-1]*f[j-i];
           }
       }
   
       num = f[n];
       free(f);
   
       return num;
   }