LeetCode:Unique Binary Search Trees I II

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

分析：依次把每一個節點做爲根節點，左邊節點做爲左子樹，右邊節點做爲右子樹，那麼總的數目等於左子樹數目*右子樹數目，實際只要求出前半部分節點做爲根節點的樹的數目，而後乘以2（奇數個節點還要加上中間節點做爲根的二叉樹數目）

遞歸代碼：爲了不重複計算子問題，用數組保存已經計算好的結果

 1 class Solution {
 2 public:
 3     int numTrees(int n) {
 4         // IMPORTANT: Please reset any member data you declared, as
 5         // the same Solution instance will be reused for each test case.
 6         int nums[n+1]; //nums[i]表示i個節點的二叉查找樹的數目
 7         memset(nums, 0, sizeof(nums));
 8         return numTreesRecur(n, nums);
 9     }
10     int numTreesRecur(int n, int nums[])
11     {
12         if(nums[n] != 0)return nums[n];
13         if(n == 0){nums[0] = 1; return 1;}
14         int tmp = (n>>1);
15         for(int i = 1; i <= tmp; i++)
16         {
17             int left,right;
18             if(nums[i-1])left = nums[i-1];
19             else left = numTreesRecur(i-1, nums);
20             if(nums[n-i])right = nums[n-i];
21             else right = numTreesRecur(n-i, nums);
22             nums[n] += left*right;
23         }
24         nums[n] <<= 1;
25         if(n % 2 != 0)
26         {
27             int val;
28             if(nums[tmp])val = nums[tmp];
29             else val = numTreesRecur(tmp, nums);
30             nums[n] += val*val;
31         }
32         return nums[n];
33     }
34 };

非遞歸代碼：從0個節點的二叉查找樹數目開始自底向上計算，dp方程爲nums[i] = sum(nums[k-1]*nums[i-k]) (k = 1,2,3...i)

 1 class Solution {
 2 public:
 3     int numTrees(int n) {
 4         // IMPORTANT: Please reset any member data you declared, as
 5         // the same Solution instance will be reused for each test case.
 6         int nums[n+1]; //num[i]表示i個節點的二叉查找樹數目
 7         memset(nums, 0, sizeof(nums));
 8         nums[0] = 1;
 9         for(int i = 1; i <= n; i++)
10         {
11             int tmp = (i>>1);
12             for(int j = 1; j <= tmp; j++)
13                 nums[i] += nums[j-1]*nums[i-j];
14             nums[i] <<= 1;
15             if(i % 2 != 0)
16                 nums[i] += nums[tmp]*nums[tmp];
17         }
18         return nums[n];
19     }
20 };

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LeetCode: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

按照上一題的思路，咱們不單單要保存i個節點對應的BST樹的數目，還要保存全部的BST樹，並且一、二、3和四、五、6雖然對應的BST數目和結構同樣，可是BST樹是不同的，由於節點值不一樣。

咱們用數組btrees[i][j][]保存節點i, i+1,...j-1,j構成的全部二叉樹，從節點數目爲1的的二叉樹開始自底向上最後求得節點數目爲n的全部二叉樹                                                                               本文地址

 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         // IMPORTANT: Please reset any member data you declared, as
14         // the same Solution instance will be reused for each test case.
15         vector<vector<vector<TreeNode*> > > btrees(n+2, vector<vector<TreeNode*> >(n+2, vector<TreeNode*>()));
16         for(int i = 1; i <= n+1; i++)
17             btrees[i][i-1].push_back(NULL); //爲了下面處理btrees[i][j]時 i > j的邊界狀況
18         for(int k = 1; k <= n; k++)//k表示節點數目
19             for(int i = 1; i <= n-k+1; i++)//i表示起始節點
20             {
21                 for(int rootval = i; rootval <= k+i-1; rootval++)
22                 {//求[i,i+1,...i+k-1]序列對應的全部BST樹
23                     for(int m = 0; m < btrees[i][rootval-1].size(); m++)//左子樹
24                         for(int n = 0; n < btrees[rootval+1][k+i-1].size(); n++)//右子樹
25                         {
26                             TreeNode *root = new TreeNode(rootval);
27                             root->left = btrees[i][rootval-1][m];
28                             root->right = btrees[rootval+1][k+i-1][n];
29                             btrees[i][k+i-1].push_back(root);
30                         }
31                 }
32             }
33         return btrees[1][n];
34     }
35 };

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