【LeetCode 中等题】50-不同的二叉搜索树II

题目描述：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

解法1。划分左右子树，递归构造。

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

class Solution(object):
    def generateTrees(self, n):
        """
        :type n: int
        :rtype: List[TreeNode]
        """
        if n == 0:
            return []
        return self.dfs(1, n)
    def dfs(self, start, end):
        if start>end:
            return [None]
        res = []
        for rootVal in range(start, end+1):
            leftTree = self.dfs(start, rootVal-1)
            rightTree = self.dfs(rootVal+1, end)
            for i in leftTree:
                for j in rightTree:
                    node = TreeNode(rootVal)
                    node.left = i
                    node.right = j
                    res.append(node)
        return res

解法2。用记忆搜索改良，用一个dict存储计算过的（start， end）的结果，避免重复计算

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

class Solution(object):
    def generateTrees(self, n):
        """
        :type n: int
        :rtype: List[TreeNode]
        """
        if n == 0:
            return []
        memo = {}
        self.dfs(1, n, memo)
        return memo[(1, n)]
    
    def dfs(self, start, end, memo):
        if start>end:
            return [None]
        if (start, end) in memo:
            return memo[(start,end)]
        res = []
        for rootVal in range(start, end+1):
            leftTree = self.dfs(start, rootVal-1, memo)
            rightTree = self.dfs(rootVal+1, end, memo)
            for i in leftTree:
                for j in rightTree:
                    node = TreeNode(rootVal)
                    node.left = i
                    node.right = j
                    res.append(node)
        memo[(start, end)] = res
        return res    # 这个地方一定要有返回值，不然上面的leftTree和rightTree会得不到返回值

参考链接：https://blog.csdn.net/qq_17550379/article/details/84635060