LeetCode 95: 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
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; left = null; right = null; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        // Use a nested for-loop to go through every possible combinations of left tree and right tree for a given root.
        // Do it recursively because it’s the same for the left tree and right tree of root.
        return genTrees(1, n);
    }
    
    private List<TreeNode> genTrees(int start, int end)
    {
        List<TreeNode> list = new LinkedList<>();
        
        // Return null branch
        if (start > end)
        {
            list.add(null);
            return list;
        }
        
        // Root node traversal from #start to #end    
        for (int i = start; i <= end; i++)
        {
            // Get all of the possible left branches
            List<TreeNode> lefts = genTrees(start, i - 1);
            
            // Get all of the possible right branches
            List<TreeNode> rights = genTrees(i + 1, end);
            
            // Nested for-loop is similar to "Unique Binary Search Trees"
            // The combinations is combination(left)*combination(right)
            for (TreeNode left : lefts)
            {
                for (TreeNode right : rights)
                {
                    TreeNode node = new TreeNode(i);
                    node.left = left;
                    node.right = right;
                    list.add(node);
                }
            }
        }
        
        return list;
    }    
}