95. 不同的二叉搜索树 II (二叉搜索树，递归，中等)

题目链接

给你一个整数 n ，请你生成并返回所有由 n 个节点组成且节点值从 1 到 n 互不相同的不同二叉搜索树。可以按任意顺序返回答案。

示例 1：

输入：n = 3
输出：[[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]

示例 2：

输入：n = 1
输出：[[1]]

提示：

• 1 <= n <= 8

解答

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        return generateBSTs(1, n);
    }

    public List<TreeNode> generateBSTs(int start, int end) {
        List<TreeNode> treeList = new ArrayList<>();
        if (start > end) return treeList;
        // 以1-n进行遍历，对原数组进行分隔
        for (int i = start; i <= end; i++) {
            List<TreeNode> leftTrees = generateBSTs(start, i-1);
            List<TreeNode> rightTrees = generateBSTs(i+1, end);
            if (leftTrees.isEmpty()) {
                if (rightTrees.isEmpty()) treeList.add(new TreeNode(i));
                else
                    for (TreeNode right : rightTrees)
                        treeList.add(new TreeNode(i, null, right));
            } else if (rightTrees.isEmpty()) {
                for (TreeNode left : leftTrees)
                    treeList.add(new TreeNode(i, left, null));
            }
            for (TreeNode left : leftTrees)
                for (TreeNode right : rightTrees)
                    treeList.add(new TreeNode(i, left, right));
        }
        return treeList;
    }
}