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
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int *f; TreeNode* generateTree(int rank, int base, int n) { if (n == 0) return NULL; if (n == 1) return new TreeNode(base); for (int i = 0; i < n; ++i) { rank -= f[i] * f[n - 1 - i]; if (rank < 0) { TreeNode* root = new TreeNode(base + i); rank += f[i] * f[n - 1 - i]; root->left = generateTree(rank % f[i], base, i); root->right = generateTree(rank / f[i], base + i + 1, n - 1 - i); return root; } } } vector<TreeNode *> generateTrees(int n) { f = new int[n + 1]; f[0] = f[1] = 1; for (int i = 2; i <= n; ++i) { f[i] = 0; for (int j = 0; j < i; ++j) f[i] += f[j] * f[i - 1 - j]; } vector<TreeNode *> result; for (int i = 0; i < f[n]; ++i) result.push_back(generateTree(i, 1, n)); return result; } };
原文:http://www.cnblogs.com/code-swan/p/4141186.html