95. Unique Binary Search Trees II java solutions

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

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该题比第一题难在要返回所有的BST树，第一题只需要返回BST树的个数

具体做法是：

1. 选出根结点后应该先分别求解该根的左右子树集合，也就是根的左子树有若干种，它们组成左子树集合，根的右子树有若干种，它们组成右子树集合。
2. 然后将左右子树相互配对，每一个左子树都与所有右子树匹配，每一个右子树都与所有的左子树匹配。然后将两个子树插在根结点上。
3. 最后，把根结点放入链表中。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<TreeNode> generateTrees(int n) {
12         if(n < 1) return new ArrayList<TreeNode>();
13         return generate(1,n);
14     }
15     
16     public List<TreeNode> generate(int left,int right){
17         List<TreeNode> ans = new ArrayList<TreeNode>();
18         if(left > right){
19             ans.add(null);
20             return ans;
21         }
22         for(int i = left; i <= right; i++){
23             List<TreeNode> leftlist = generate(left,i-1);
24             List<TreeNode> rightlist = generate(i+1,right);
25             for(TreeNode le : leftlist){
26                 for(TreeNode ri : rightlist){
27                     TreeNode node = new TreeNode(i);
28                     node.left = le;
29                     node.right = ri;
30                     ans.add(node);
31                 }
32             }
33         }
34         return ans;
35     }
36 }

第一题链接：

96. Unique Binary Search Trees java solutions