leetcode--95. Unique Binary Search Trees II

1、原题描述

95. Unique Binary Search Trees II

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

2、思路：II与I是一样的题目只是返回结果不同，I只需要返回可以生成的BST的数目，II要求返回所有Tree。II的难点在于怎么返回这个结果。基本代码思路还是I，[st, i - 1]和[i+1, end]

   返回以此为范围生成的所有tree，然后两者合并，当然合并时需要把每棵树都要copy一份。

3、代码实现

/**
 * 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 {
public:
    vector<TreeNode*> generateTrees(int n) {
        return generate_bst(1, n);
    }
    
    vector<TreeNode*> generate_bst(int st, int end) {
        vector<TreeNode*> results;
        for (int i = st; i <= end; i++) {
            vector<TreeNode*> res_left = generate_bst(st, i - 1);
            vector<TreeNode*> res_right = generate_bst(i + 1, end);
            save_results(results, i, res_left, res_right);
        }
        return results;
    }
    
    void save_results(vector<TreeNode*>& results, const int root_val, 
                      const vector<TreeNode*>& res_left, const vector<TreeNode*>& res_right) {
        int num_left = res_left.size();
        int num_right = res_right.size();
        if (num_left == 0 && num_right == 0) {
            TreeNode* root = new TreeNode(root_val);
            return results.push_back(root);
        } else if (num_left == 0) {
            for (int i = 0; i < num_right; i++) {
                TreeNode* root = new TreeNode(root_val);
                root->right = copy_tree(res_right[i]);
                results.push_back(root);
            }
        } else if (num_right == 0) {
            for (int i = 0; i < num_left; i++) {
                TreeNode* root = new TreeNode(root_val);
                root->left = copy_tree(res_left[i]);
                results.push_back(root);
            }
        } else {
            for (int i = 0; i < num_left; i++) {
                for (int j = 0; j < num_right; j++) {
                    TreeNode* root = new TreeNode(root_val);
                    root->left = copy_tree(res_left[i]);
                    root->right = copy_tree(res_right[j]);
                    results.push_back(root);
                }
            }
        }
        return;
    }
    
    TreeNode* copy_tree(TreeNode* root) {
        if (root == NULL) {
            return NULL;
        }
        
        TreeNode* node = new TreeNode(root->val);
        node->left = copy_tree(root->left);
        node->right = copy_tree(root->right);
        return node;
    }
};

4、优化

   应该是没有优化空间了，返回结果决定了解空间，后续看看在一些小操作上面有没有优化空间。

5、api

verctor： push_back只能在最后追加(添加)一个元素。似乎没有方法能一次添加多个元素，除了初始化时。