Boostable

  博客园  :: 首页  :: 新随笔  :: 联系 :: 订阅 订阅  :: 管理

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 };

 



posted on 2014-08-31 21:43  Boostable  阅读(221)  评论(0编辑  收藏  举报