Unique Binary Search Trees II



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:
    vector<TreeNode *> generateTrees(int n) {
        vector<TreeNode *> allTrees = generateTrees(1,n);
        return allTrees;
    }
    
private:
   vector<TreeNode *> generateTrees(int low,int high){
        vector<TreeNode *> allTrees;
        if(low > high) {
            allTrees.push_back(NULL);
            return allTrees;
        }
        if(low == high){
            TreeNode *p = new TreeNode(low);
            allTrees.push_back(p);
            return allTrees;
        }
        for(int i = low ;i <= high ; ++i){
            vector<TreeNode *> leftTrees = generateTrees(low,i-1);//把i当做根节点的值，leftTrees表示所有可能的左子树根节点
            vector<TreeNode *> rightTrees = generateTrees(i+1,high);//把i当做根节点的值，rightTrees表示所有可能的右子树根节点
            for(int j=0;j<leftTrees.size();++j){
                for(int k=0;k<rightTrees.size();++k){
                    TreeNode* root = new TreeNode(i);//i作为根节点的值
                    root->left = leftTrees[j];
                    root->right = rightTrees[k];
                    allTrees.push_back(root);
                }
            }
        }
        return allTrees;
   }
};