Unique_Binary_Search_Trees_II.txt

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 {
public:
    void generateTreesHelper(int start,int end, vector<TreeNode *> &res){
        if (start>end){
            res.push_back(NULL);
            return;
        }
        for (int i=start; i<=end; ++i){
            vector<TreeNode *> leftres,rightres;
            generateTreesHelper(start,i-1,leftres);
            generateTreesHelper(i+1,end,rightres);
            for (int j=0; j<leftres.size(); ++j){
                for (int k=0; k<rightres.size(); ++k){
                    TreeNode *root = new TreeNode(i);
                    root->left = leftres[j];
                    root->right = rightres[k];
                    res.push_back(root);
                }
            }
        }   
        
    }

    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *> res;
        generateTreesHelper(1,n,res);
        return res;
    }
};
REMARK:
1.Hard question. First of all, we need to figure out what the output is like. The output is a vector(or string) of type TreeNode*. Each element in the vector is a root corresponding to different structurally unique BST. Then, similar to the problem "Unique Binary Search Trees", i loop through the root of the tree. Next, we store the trees into leftres and rightres. Finally we assign the left child and right child of root using j and k.