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面试题22:有序数组生成BST
- 对于一个含有n个数的有序数组1~N,能够产生多少种不同结果的二叉搜素树BST?
- 如何生成这些不同结构的BST?
1 class Solution { 2 public: 3 int numTrees(int n) { 4 int* dp = new int[n+1]; 5 dp[0] = 1; 6 dp[1] = 1; 7 dp[2] = 2; 8 for(int i=3;i<=n;i++){ 9 dp[i] = 0; 10 for(int j=1;j<=i;j++){ 11 dp[i]+=dp[j-1]*dp[i-j]; 12 } 13 } 14 return dp[n]; 15 } 16 17 int numTrees2(int n){ 18 if(n==0) return 1; 19 if(n < 3) return n; 20 int sum = 0; 21 for(int i=1;i<=n;i++){ 22 sum += numTrees2(i-1)*numTrees2(n-i); 23 } 24 return sum; 25 } 26 };
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 11 class Solution { 12 public: 13 vector<TreeNode *> generateTrees(int left, int right) { 14 vector<TreeNode*> res; 15 if (left > right) { 16 res.push_back(nullptr); 17 return res; 18 } 19 20 if(left == right){ 21 res.push_back(new TreeNode(left)); 22 return res; 23 } 24 25 for (int i = left; i <= right; i++) { 26 //TreeNode* root = new TreeNode(i); 27 vector<TreeNode*> leftRes = generateTrees(left, i-1); 28 vector<TreeNode*> rightRes = generateTrees(i + 1, right); 29 for (size_t j = 0; j < leftRes.size(); j++) { 30 for (size_t k = 0; k < rightRes.size(); k++) { 31 TreeNode* root = new TreeNode(i); 32 root->left = leftRes[j]; 33 root->right = rightRes[k]; 34 res.push_back(root); 35 } 36 } 37 } 38 return res; 39 } 40 41 vector<TreeNode*> generateTrees(int n){ 42 vector<TreeNode*> res; 43 if(n == 0) { 44 res.push_back(nullptr); 45 return res; 46 } 47 return generateTrees(1,n); 48 } 49 50 };
面试题22:有序数组生成BST
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