题目：

给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。

示例:

输入: 3
输出:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
解释:
以上的输出对应以下 5 种不同结构的二叉搜索树：

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

解题：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {public IList<TreeNode> GenerateTrees(int n)
        {
            var preResult = new List<TreeNode>();
            if (n == 0)
            {
                return preResult;
            }

            preResult.Add(null);

            for (int i = 1; i <= n; i++)
            {
                var currentResult = new List<TreeNode>();

                foreach (var node in preResult)
                {
                    //放在顶点
                    var newHeadNode = new TreeNode(i)
                    {
                        left = TreeCopy(node)
                    };
                    currentResult.Add(newHeadNode);

                    //放在上一次结果的右顶点
                    //count用来记录下一次遍历到哪个顶点
                    int count = 1;
                    while (count > 0)
                    {
                        var newTreeNode = TreeCopy(node);
                        var targetNode = newTreeNode;
                        var newRigthNode = new TreeNode(i);

                        for (int j = 1; j < count; j++)
                        {
                            targetNode = targetNode.right;
                        }
                        if (targetNode != null)
                        {
                            var tempNode = targetNode.right;
                            targetNode.right = newRigthNode;
                            newRigthNode.left = tempNode;
                            currentResult.Add(newTreeNode);
                            count++;
                        }
                        else
                        {
                            count = -1;
                        }

                    }

                    preResult = currentResult;
                }
            }

            return preResult;
        }

        private TreeNode TreeCopy(TreeNode treeNode)
        {
            if (treeNode == null)
            {
                return null;
            }
            var result = new TreeNode(treeNode.val)
            {
                left = TreeCopy(treeNode.left),
                right = TreeCopy(treeNode.right)
            };
            return result;
        }
}

来源：

力扣（LeetCode）