【发布时间】:2016-03-16 01:09:59
【问题描述】:
这里我试图将二叉树组合成表达式。比如
b1 = 二叉树(3.0)
打印(括号(b1))
3.0
b2 = 二叉树(4.0)
b3 = 二叉树(7.0)
b4 = BinaryTree("*", b1, b2)
b5 = BinaryTree("+", b4, b3)
打印(括号(b5))`
((3.0 * 4.0) + 7.0)
我为parenthesize() 编写的代码在底部,上面是二叉树基础代码。但是我的代码不是返回“somenumber”,而是返回
(“某个数字”)。当我删除“(”时,括号完全消失了。有人可以帮忙填写代码吗?
class BinaryTree:
"""
A Binary Tree, i.e. arity 2.
=== Attributes ===
@param object data: data for this binary tree node
@param BinaryTree|None left: left child of this binary tree node
@param BinaryTree|None right: right child of this binary tree node
"""
def __init__(self, data, left=None, right=None):
"""
Create BinaryTree self with data and children left and right.
@param BinaryTree self: this binary tree
@param object data: data of this node
@param BinaryTree|None left: left child
@param BinaryTree|None right: right child
@rtype: None
"""
self.data, self.left, self.right = data, left, right
def __eq__(self, other):
"""
Return whether BinaryTree self is equivalent to other.
@param BinaryTree self: this binary tree
@param Any other: object to check equivalence to self
@rtype: bool
>>> BinaryTree(7).__eq__("seven")
False
>>> b1 = BinaryTree(7, BinaryTree(5))
>>> b1.__eq__(BinaryTree(7, BinaryTree(5), None))
True
"""
return (type(self) == type(other) and
self.data == other.data and
(self.left, self.right) == (other.left, other.right))
def __repr__(self):
"""
Represent BinaryTree (self) as a string that can be evaluated to
produce an equivalent BinaryTree.
@param BinaryTree self: this binary tree
@rtype: str
>>> BinaryTree(1, BinaryTree(2), BinaryTree(3))
BinaryTree(1, BinaryTree(2, None, None), BinaryTree(3, None, None))
"""
return "BinaryTree({}, {}, {})".format(repr(self.data),
repr(self.left),
repr(self.right))
def __str__(self, indent=""):
"""
Return a user-friendly string representing BinaryTree (self)
inorder. Indent by indent.
>>> b = BinaryTree(1, BinaryTree(2, BinaryTree(3)), BinaryTree(4))
>>> print(b)
4
1
2
3
<BLANKLINE>
"""
right_tree = (self.right.__str__(
indent + " ") if self.right else "")
left_tree = self.left.__str__(indent + " ") if self.left else ""
return (right_tree + "{}{}\n".format(indent, str(self.data)) +
left_tree)
def __contains__(self, value):
"""
Return whether tree rooted at node contains value.
@param BinaryTree self: binary tree to search for value
@param object value: value to search for
@rtype: bool
>>> BinaryTree(5, BinaryTree(7), BinaryTree(9)).__contains__(7)
True
"""
return (self.data == value or
(self.left and value in self.left) or
(self.right and value in self.right))
def parenthesize(b):
"""
Return a parenthesized expression equivalent to the arithmetic
expression tree rooted at b.
Assume: -- b is a binary tree
-- interior nodes contain data in {'+', '-', '*', '/'}
-- interior nodes always have two children
-- leaves contain float data
@param BinaryTree b: arithmetic expression tree
@rtype: str
>>> b1 = BinaryTree(3.0)
>>> print(parenthesize(b1))
3.0
>>> b2 = BinaryTree(4.0)
>>> b3 = BinaryTree(7.0)
>>> b4 = BinaryTree("*", b1, b2)
>>> b5 = BinaryTree("+", b4, b3)
>>> print(parenthesize(b5))
((3.0 * 4.0) + 7.0)
"""
if b is None:
return ''
else:
return "("+parenthesize(b.left)+str(b.data)+parenthesize(b.right)+")"
【问题讨论】:
标签: python python-3.x tree binary-tree