leetcode python3 不同的二叉搜索树

代码思路:利用递归思想,遍历1到n的数字i作为根节点,i左侧的点为该树的左节点,右侧的点为右节点,递归得到所有左子树和右子树后将结果输出

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def generateTrees(self, n: int) -> List[TreeNode]:
        def generateTree(start,end):
            if start>end:
                return [None,]
            all_trees=[]
            for i in range(start, end + 1):  # pick up a root
                leftTrees=generateTree(start,i-1)
                rightTrees=generateTree(i+1,end)
                for l in leftTrees:
                    for r in rightTrees:
                        current_tree = TreeNode(i)
                        current_tree.left = l
                        current_tree.right = r
                        all_trees.append(current_tree)
            return all_trees

        return generateTree(1, n) if n else []
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转载自blog.csdn.net/m0_37656366/article/details/105022377