95. Unique Binary Search Trees II (Medium)

Given an integer 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

Solution: DFS

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<TreeNode*> dfs(int start, int end) {
        if (start > end) return vector<TreeNode*>(1,NULL);
        vector<TreeNode*> subTree;
        for (int i = start; i <= end; ++i) {
            vector<TreeNode*> leftSubTree = dfs(start,i-1);
            vector<TreeNode*> rightSubTree = dfs(i+1,end);
            for (int j = 0; j < leftSubTree.size(); ++j) {
                for (int k = 0; k < rightSubTree.size(); ++k) {
                    TreeNode* root = new TreeNode(i);
                    root->left = leftSubTree[j];
                    root->right = rightSubTree[k];
                    subTree.push_back(root);
                }
            }
        }
        return subTree;
    }
public:
    vector<TreeNode*> generateTrees(int n) {
        if (n == 0) return {};
        return dfs(1,n);
    }
};