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
;
}
};

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;
}
};