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 。
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
0 <= n <= 8
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def generateTrees(self, n: int) -> List[TreeNode]:
def helper(m: int, n: int) -> List[TreeNode]:
if m > n:
return [None]
ret = []
for val in range(m, n + 1):
for left in helper(m, val - 1):
for right in helper(val + 1, n):
ret.append(TreeNode(val, left, right))
return ret
return helper(1, n) if n > 0 else []