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：依次把每个节点作为根节点，左边节点作为左子树，右边节点作为右子树，那么总的数目等于左子树数目*右子树数目，实际只要求出前半部分节点作为根节点的树的数目，然后乘以2（奇数个节点还要加上中间节点作为根的二叉树数目）

递归代码：为了避免重复计算子问题，用数组保存已经计算好的结果

 class Solution {
 public:
     int numTrees(int n) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         int nums[n+]; //nums[i]表示i个节点的二叉查找树的数目
         memset(nums, , sizeof(nums));
         return numTreesRecur(n, nums);
     }
     int numTreesRecur(int n, int nums[])
     {
         if(nums[n] != )return nums[n];
         if(n == ){nums[] = ; return ;}
         int tmp = (n>>);
         for(int i = ; i <= tmp; i++)
         {
             int left,right;
             if(nums[i-])left = nums[i-];
             else left = numTreesRecur(i-, nums);
             if(nums[n-i])right = nums[n-i];
             else right = numTreesRecur(n-i, nums);
             nums[n] += left*right;
         }
         nums[n] <<= ;
         if(n %  != )
         {
             int val;
             if(nums[tmp])val = nums[tmp];
             else val = numTreesRecur(tmp, nums);
             nums[n] += val*val;
         }
         return nums[n];
     }
 };

非递归代码：从0个节点的二叉查找树数目开始自底向上计算，dp方程为nums[i] = sum(nums[k-1]*nums[i-k]) (k = 1,2,3...i)

 class Solution {
 public:
     int numTrees(int n) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         int nums[n+]; //num[i]表示i个节点的二叉查找树数目
         memset(nums, , sizeof(nums));
         nums[] = ;
         for(int i = ; i <= n; i++)
         {
             int tmp = (i>>);
             for(int j = ; j <= tmp; j++)
                 nums[i] += nums[j-]*nums[i-j];
             nums[i] <<= ;
             if(i %  != )
                 nums[i] += nums[tmp]*nums[tmp];
         }
         return nums[n];
     }
 };

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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

按照上一题的思路，我们不仅仅要保存i个节点对应的BST树的数目，还要保存所有的BST树，而且1、2、3和4、5、6虽然对应的BST数目和结构一样，但是BST树是不一样的，因为节点值不同。

我们用数组btrees[i][j][]保存节点i, i+1,...j-1,j构成的所有二叉树，从节点数目为1的的二叉树开始自底向上最后求得节点数目为n的所有二叉树                                                                               本文地址

 /**
  * Definition for binary tree
  * 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) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<vector<vector<TreeNode*> > > btrees(n+, vector<vector<TreeNode*> >(n+, vector<TreeNode*>()));
         for(int i = ; i <= n+; i++)
             btrees[i][i-].push_back(NULL); //为了下面处理btrees[i][j]时 i > j的边界情况
         for(int k = ; k <= n; k++)//k表示节点数目
             for(int i = ; i <= n-k+; i++)//i表示起始节点
             {
                 for(int rootval = i; rootval <= k+i-; rootval++)
                 {//求[i,i+1,...i+k-1]序列对应的所有BST树
                     for(int m = ; m < btrees[i][rootval-].size(); m++)//左子树
                         for(int n = ; n < btrees[rootval+][k+i-].size(); n++)//右子树
                         {
                             TreeNode *root = new TreeNode(rootval);
                             root->left = btrees[i][rootval-][m];
                             root->right = btrees[rootval+][k+i-][n];
                             btrees[i][k+i-].push_back(root);
                         }
                 }
             }
         return btrees[][n];
     }
 };

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