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

如果F(n)表示长度为n的二叉树有多少种结果，则

F(n) = F(0)*F(n-1) + F(1)*F(n-2) + F(2)*F(n-2) + ......+ F(n-1)*F(0)

所以代码如下：

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

Unique Binary Search Trees II

Given n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

For example,

Given n = 3, your program should return all 5 unique BST’s shown below.

1         3     3      2      1    \       /     /      / \      \     3     2     1      1   3      2    /     /       \                 \   2     1         2                 3

同样沿用 I 的思路，利用分治思想

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector
     
       generateTrees(int n) {        if(n <=0) return vector
      
       ();        return generateSubTrees(1,n);    }    vector
       
        generateSubTrees(int s,int e){        vector
        
          res;        if(s > e){            res.push_back(NULL);            return res;        }        for(int i = s; i <= e;i++ ){            vector
         
           left = generateSubTrees(s,i-1);            vector
          
            right = generateSubTrees(i+1,e); for(auto l : left){ for(auto r : right){ TreeNode* root = new TreeNode(i); root->left = l; root->right = r; res.push_back(root); } } } return res; }};