unique-binary-search-trees Java code

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.
OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; left = null; right = null; } * } */import java.util.*;public class Solution {    public ArrayList<TreeNode> generateTrees(int n) {        return createTree(1, n);    }    public ArrayList<TreeNode> createTree(int low, int high) {        ArrayList<TreeNode> result = new ArrayList<>();        if(low > high) {            result.add(null);            return result;        }        for (int i = low; i <= high; i ++) {            // 求根结点i的左右子树集合            ArrayList<TreeNode> left = createTree(low, i - 1);            ArrayList<TreeNode> right = createTree(i + 1, high);            for (int j = 0; j < left.size(); j ++) {                for (int k = 0; k < right.size(); k ++) {                    // 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                    TreeNode newNode = new TreeNode(i);                    newNode.left = left.get(j);                    newNode.right = right.get(k);                    result.add(newNode);                }            }        }        return result;    }}