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给定n求二叉搜索树的个数
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
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {private: vector<TreeNode *> generate(int l, int r) { vector<TreeNode *> ans; if (l > r) { ans.push_back(NULL); } else { for (int i=l; i<=r; i++){ vector<TreeNode *> left = generate(l, i-1); vector<TreeNode *> right = generate(i+1, r); for (int j=0; j<left.size(); j++) { for (int k=0; k<right.size(); k++) { TreeNode *pRoot = new TreeNode(i); pRoot->left = left[j]; pRoot->right = right[k]; ans.push_back(pRoot); } } } } return ans; }public: vector<TreeNode *> generateTrees(int n) { return generate(1, n); }};
给定n求二叉搜索树的个数
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