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
地址：https://oj.leetcode.com/problems/unique-binary-search-trees/
算法：动态规划，dp[i]记录有i个节点的搜索树数量，则$dp[i+1]=sum_{j=0}^{i-1}dp[j]*dp[i-1-j]$，代码：
1 class Solution {
2 public:
3 int numTrees(int n) {
4 vector<int> dp(n+1);
5 dp[0] = 1;
6 dp[1] = 1;
7 for(int i = 2; i <= n; ++i){
8 int sum = 0;
9 for(int j = i-1; j >= 0; --j){
10 sum += (dp[j] * dp[i-1-j]);
11 }
12 dp[i] = sum;
13 }
14 return dp[n];
15 }
16 };
第二题：
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
地址：https://oj.leetcode.com/problems/unique-binary-search-trees-ii/
算法：与上一题不同，本题要求构造出所有的解。用递归完成的，代码：
1 /**
2 * Definition for binary tree
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 vector<TreeNode *> generateTrees(int n) {
13 TreeNode *p = NULL;
14 if(n <= 0) return vector<TreeNode*>(1,p);
15 return generateSubtrees(1,n);
16 }
17 vector<TreeNode *> generateSubtrees(int s, int e){
18 TreeNode *p = NULL;
19 if(s > e) return vector<TreeNode*>(1,p);
20 if(s == e) {
21 p = new TreeNode(s);
22 return vector<TreeNode*>(1,p);
23 }
24 vector<TreeNode*> result;
25 for(int i = s; i <= e; ++i){
26 vector<TreeNode*> left = generateSubtrees(s,i-1);
27 vector<TreeNode*> right = generateSubtrees(i+1,e);
28 for(int left_i = 0; left_i < left.size(); ++left_i)
29 for(int right_i = 0; right_i < right.size(); ++right_i){
30 p = new TreeNode(i);
31 p->left = left[left_i];
32 p->right = right[right_i];
33 result.push_back(p);
34 }
35 }
36 return result;
37 }
38 };