Question

我试过差不多了，我通过了第一个测试用例而不是最后两个测试用例这对大树的正确性和性能是错误的，它必须返回True或False。

import collections


class BinarySearchTree:
      Node = collections.namedtuple('Node', ['left', 'right', 'value'])

      @staticmethod
      def contains(root, value):
          if root.value == value:
              return True
          elif root.value < value:
               return root.right.value == value #I am using == to return True or  
                                         #False
          else:
              return root.right.value == value


n1 = BinarySearchTree.Node(value=1, left=None, right=None)
n3 = BinarySearchTree.Node(value=3, left=None, right=None)
n2 = BinarySearchTree.Node(value=2, left=n1, right=n3)


print(BinarySearchTree.contains(n2, 3))

如何遍历元组中的元组内的元组？

Answer 1

试试这段代码。

以前的代码是针对一个节点而不是整个树。

else也检查了正确的节点而不是左边的节点。

import collections


class BinarySearchTree:
      Node = collections.namedtuple('Node', ['left', 'right', 'value'])


  @staticmethod
  def contains(root, value):
      if root.value == value:
          return True
      elif root.value < value:
          if root.right == None:
               return False
          else:
               return BinarySearchTree.contains(root.right,value)
        #root.right.value == value #I am using == to return True or  
                                     #False
      else:
          if root.left == None:
              return False
          else:
              return BinarySearchTree.contains(root.left,value)
          #root.right.value == value


n4 = BinarySearchTree.Node(value=4, left=None, right=None)
n0 = BinarySearchTree.Node(value=0, left=None, right=None)
n1 = BinarySearchTree.Node(value=1, left=n0, right=None)
n3 = BinarySearchTree.Node(value=3, left=None, right=n4)
n2 = BinarySearchTree.Node(value=2, left=n1, right=n3)


print(BinarySearchTree.contains(n2, 14))

编写一个函数，使用python

1 个答案: