Description:

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

和第I题的区别是这题要返回所有的结果，不是结果的数目。所以要递归枚举：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        public List
     
       generate(int start, int end) {        List
      
        ret = new ArrayList
       
        ();                if(start > end) {            ret.add(null);            return ret;        }                for(int i=start; i<=end; i++) {            List
        
          lTree = generate(start, i-1); //生成左子树集合            List
         
           rTree = generate(i+1, end);  //生成右子树集合            for(int j=0; j
          
            generateTrees(int n) { if(n <= 0) return new ArrayList
           
            (); return generate(0, n-1); }}
           
          
         
        
       
      
     

C++:

/** * 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
     
       generate(int start, int end)    {        vector
      
        ret;        if (start > end)        {            ret.push_back(NULL);            return ret;        }                for(int i = start; i <= end; i++)        {            vector
       
         lTree = generate(start, i - 1);            vector
        
          rTree = generate(i + 1, end);            for(int j = 0; j < lTree.size(); j++)                for(int k = 0; k < rTree.size(); k++)                {                    TreeNode *node = new TreeNode(i + 1);                    ret.push_back(node);                    node->left = lTree[j];                    node->right = rTree[k];                              }                   }                return ret;    }        vector
         
           generateTrees(int n) {
            vector
          
            ret; if(n <= 0) return ret; return generate(0, n - 1); }};
          
         
        
       
      
     

同样的代码同样的测试数据Java代码为什么效率高这么多，难道是跑代码的服务器不一样？还是Java后台调用代码写的好？