[笔记]算法学习:不同的二叉搜索树 II

zerglurker

题目:

https://leetcode-cn.com/problems/unique-binary-search-trees-ii/description/

解题要点:1 树类型的算法,一般都可以用递归来解决,因此首先考虑递归2 接着,要考虑如何生成树3 最后考虑如何输出树具体内容看下面代码注释import json

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import json     # Definition for a binary tree node. class TreeNode:  # 树的数据结构     def __init__(self, x):         self.val = x         self.left = None         self.right = None     class Solution:     def generateTrees(self, n):         """         :type n: int         :rtype: List[TreeNode]         """         if n == 0:  # 特殊情况处理             return []         if n == 1:  # 特殊情况处理             return [[1]]         total = [x for x in range(1, n + 1)]  # 首先生成点的集合         nodes = self.makeNode(total)  # 生成树         result = list()         for node in nodes:             d = self.dumpTreeNode(node)  # 导出结果树集合             tmp = list()             for i in range(9999):  # 一层一层的将树输入到列表                 if i not in d:                     break                 tmp.extend(d[i])             while tmp[-1] is None:  # 将尾部的空节点删除                 del tmp[-1]             result.append(tmp)  # 将列表加入结果集         return result       def makeNode(self, item_set: list) -> list:  # 生成树         if len(item_set) == 0:  # 节点集没有一个点             return [None]         if len(item_set) == 1:  # 节点集只有一个点             return [TreeNode(item_set[0])]         result = list()         if len(item_set) == 2:  # 节点集只有两个点             if item_set[0] < item_set[1]:                 node = TreeNode(item_set[0])                 node.right = TreeNode(item_set[1])                 result.append(node)                 node = TreeNode(item_set[1])                 node.left = TreeNode(item_set[0])                 result.append(node)             else:                 node = TreeNode(item_set[0])                 node.left = TreeNode(item_set[1])                 result.append(node)                 node = TreeNode(item_set[1])                 node.right = TreeNode(item_set[0])                 result.append(node)             return result         for i in item_set:  # 节点集有多个点             left = list()             right = list()             for j in item_set:  # 遍历所有点,依次做根节点                 if j == i:                     continue                 if j < i:                     left.append(j)                     continue                 right.append(j)             ll = self.makeNode(left)             rr = self.makeNode(right)             for l_ in ll:                 for r_ in rr:  # 排列组合左右子树                     node = TreeNode(i)                     node.left = l_                     if r_ is not None:                         node.right = r_                     result.append(node)         return result       def dumpTreeNode(self, node: TreeNode, level: int = 0) -> dict:  # 导出树         result = dict()         result[level] = [node.val]         if isinstance(node.left, TreeNode):             r = self.dumpTreeNode(node.left, level + 1)  # 递归导出子树             for k in r:                 if k not in result:                     result[k] = list()                 result[k].extend(r[k])         else:             if level + 1 not in result:                 result[level + 1] = list()             result[level + 1].append(node.left)         if isinstance(node.right, TreeNode):             r = self.dumpTreeNode(node.right, level + 1)  # 递归导出子树             for k in r:                 if k not in result:                     result[k] = list()                 result[k].extend(r[k])         else:             if level + 1 not in result:                 result[level + 1] = list()             result[level + 1].append(node.right)         return result     def main():  # 测试函数主函数     test = [         '[]',         '[[1]]',         '[[1,null,2],[2,1]]',         '[[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]',         '[[1,null,2,null,3,null,4],[1,null,2,null,4,3],[1,null,3,2,4],[1,null,4,2,null,null,3],[1,null,4,3,null,2],[2,1,3,null,null,null,4],[2,1,4,null,null,3],[3,1,4,null,2],[3,2,4,1],[4,1,null,null,2,null,3],[4,1,null,null,3,2],[4,2,null,1,3],[4,3,null,1,null,null,2],[4,3,null,2,null,1]]',         '[[1,null,2,null,3,null,4,null,5],[1,null,2,null,3,null,5,4],[1,null,2,null,4,3,5],[1,null,2,null,5,3,null,null,4],[1,null,2,null,5,4,null,3],[1,null,3,2,4,null,null,null,5],[1,null,3,2,5,null,null,4],[1,null,4,2,5,null,3],[1,null,4,3,5,2],[1,null,5,2,null,null,3,null,4],[1,null,5,2,null,null,4,3],[1,null,5,3,null,2,4],[1,null,5,4,null,2,null,null,3],[1,null,5,4,null,3,null,2],[2,1,3,null,null,null,4,null,5],[2,1,3,null,null,null,5,4],[2,1,4,null,null,3,5],[2,1,5,null,null,3,null,null,4],[2,1,5,null,null,4,null,3],[3,1,4,null,2,null,5],[3,1,5,null,2,4],[3,2,4,1,null,null,5],[3,2,5,1,null,4],[4,1,5,null,2,null,null,null,3],[4,1,5,null,3,null,null,2],[4,2,5,1,3],[4,3,5,1,null,null,null,null,2],[4,3,5,2,null,null,null,1],[5,1,null,null,2,null,3,null,4],[5,1,null,null,2,null,4,3],[5,1,null,null,3,2,4],[5,1,null,null,4,2,null,null,3],[5,1,null,null,4,3,null,2],[5,2,null,1,3,null,null,null,4],[5,2,null,1,4,null,null,3],[5,3,null,1,4,null,2],[5,3,null,2,4,1],[5,4,null,1,null,null,2,null,3],[5,4,null,1,null,null,3,2],[5,4,null,2,null,1,3],[5,4,null,3,null,1,null,null,2],[5,4,null,3,null,2,null,1]]'     ]     for i in range(len(test)):         s = Solution()         r = s.generateTrees(i)         s = test[i]         r_ = json.loads(s)         print(len(r), len(r_))         for j, v in enumerate(r):             if v != r_[j]:                 print(i, v == r_[j], v, r_[j])     if __name__ == '__main__':     main()

wangqiustc

python的类有点复杂，我是能避开就避开了

zerglurker

qcztsn 发表于 2018-11-16 19:51
楼主这个是leetcode上的题目吗？

是的,最近下班了就会研究这个,把自己写的代码发这里来
这个是提交通过的
上面的代码不但有正确性的限制,还有性能上面的限制
有兴趣也可以一起啊

zerglurker

wangqiustc 发表于 2018-11-16 18:28
python的类有点复杂，我是能避开就避开了

python3已经改进很多了,加入了保护成员和方法(名称前面加下划线)
python的类其实还是很容易的
属性和方法和其他相比更简单

小黑LLB

二叉树搜索有什么简单的例子么，看上去好复杂，先收藏一下，之后再研究一下