96. Unique Binary Search Trees（I 和 II）

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

https://leetcode.com/discuss/24282/dp-solution-in-6-lines-with-explanation-f-i-n-g-i-1-g-n-i

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<int> 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<TreeNode*> generateTrees(int n) {
        if(n <=0) return vector<TreeNode*>();
        return generateSubTrees(1,n);
    }
    vector<TreeNode*>generateSubTrees(int s,int e){
        vector<TreeNode*> res;
        if(s > e){
            res.push_back(NULL);
            return res;
        }
        for(int i = s; i <= e;i++ ){
            vector<TreeNode*> left = generateSubTrees(s,i-1);
            vector<TreeNode*> 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;
    }
};