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
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个节点的二叉查找树的数目 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)
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个节点的二叉查找树数目 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 };
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
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); //为了下面处理btrees[i][j]时 i > j的边界情况 18 for(int k = 1; k <= n; k++)//k表示节点数目 19 for(int i = 1; i <= n-k+1; i++)//i表示起始节点 20 { 21 for(int rootval = i; rootval <= k+i-1; rootval++) 22 {//求[i,i+1,...i+k-1]序列对应的所有BST树 23 for(int m = 0; m < btrees[i][rootval-1].size(); m++)//左子树 24 for(int n = 0; n < btrees[rootval+1][k+i-1].size(); n++)//右子树 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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