def infix_to_postfix(infix): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} postfix = '' stack = Stack() for ch in infix: if ch == '(': stack.push(ch) elif ch == ')': while stack.top() != '(': postfix += stack.pop() stack.pop() elif ch in precedence: while stack.is_empty() == False and stack.top() != '(' and precedence[ch] <= precedence[stack.top()]: postfix += stack.pop() stack.push(ch) else: postfix += ch while stack.is_empty() == False: postfix += stack.pop() return postfix解释这段代码

这段代码是一个中缀表达式转后缀表达式的实现。它使用了一个栈来辅助转换过程。具体来说，它遍历中缀表达式的每个字符，如果是左括号，就将它压入栈中；如果是右括号，则将栈中的元素弹出并加入后缀表达式中，直到遇到左括号；如果是运算符，则将栈中优先级大于等于它的元素弹出并加入后缀表达式中，然后将它自己压入栈中；如果是操作数，则直接加入后缀表达式中。最后，将栈中剩余的元素依次弹出并加入后缀表达式中，即得到了后缀表达式。

相关问题

这是上题的代码：def infix_to_postfix(expression): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} op_stack = [] postfix_list = [] token_list = expression.split() for token in token_list: if token.isalnum(): postfix_list.append(token) elif token == '(': op_stack.append(token) elif token == ')': top_token = op_stack.pop() while top_token != '(': postfix_list.append(top_token) top_token = op_stack.pop() else: # operator while op_stack and precedence[op_stack[-1]] >= precedence[token]: postfix_list.append(op_stack.pop()) op_stack.append(token) while op_stack: postfix_list.append(op_stack.pop()) return ' '.join(postfix_list) class Node: def __init__(self, value): self.value = value self.left_child = None self.right_child = None def build_expression_tree(postfix_expr): operator_stack = [] token_list = postfix_expr.split() for token in token_list: if token.isalnum(): node = Node(token) operator_stack.append(node) else: right_node = operator_stack.pop() left_node = operator_stack.pop() node = Node(token) node.left_child = left_node node.right_child = right_node operator_stack.append(node) return operator_stack.pop() def evaluate_expression_tree(node, variable_values): if node.value.isalnum(): return variable_values[node.value] else: left_value = evaluate_expression_tree(node.left_child, variable_values) right_value = evaluate_expression_tree(node.right_child, variable_values) if node.value == '!': return not left_value elif node.value == '&': return left_value and right_value elif node.value == '|': return left_value or right_value expression = "!a & (b | c)" postfix_expression = infix_to_postfix(expression) expression_tree = build_expression_tree(postfix_expression) variable_values = {'a': True, 'b': False, 'c': True} result = evaluate_expression_tree(expression_tree, variable_values) print(result)

好的，感谢提供代码。以下是针对你的问题的解答：

1. 打印二叉树构造过程

在 build_expression_tree 函数中，operator_stack 列表存储了构建二叉树的过程。我们可以在函数中添加打印语句来查看二叉树构建过程。

代码如下：

def build_expression_tree(postfix_expr):
    operator_stack = []
    token_list = postfix_expr.split()

    for token in token_list:
        if token.isalnum():
            node = Node(token)
            operator_stack.append(node)
        else:
            right_node = operator_stack.pop()
            left_node = operator_stack.pop()
            node = Node(token)
            node.left_child = left_node
            node.right_child = right_node
            operator_stack.append(node)
        
        # 打印二叉树构建过程
        print([i.value for i in operator_stack])

    return operator_stack.pop()

在运行 build_expression_tree 函数时，将会打印出二叉树构建过程。

2. 打印后缀表达式和二叉树后序遍历序列

在函数 infix_to_postfix 中，postfix_list 列表存储了后缀表达式。我们可以在函数返回后缀表达式的同时，打印出后缀表达式。

代码如下：

def infix_to_postfix(expression):
    precedence = {'!': 3, '&': 2, '|': 1, '(': 0}
    op_stack = []
    postfix_list = []
    token_list = expression.split()

    for token in token_list:
        if token.isalnum():
            postfix_list.append(token)
        elif token == '(':
            op_stack.append(token)
        elif token == ')':
            top_token = op_stack.pop()
            while top_token != '(':
                postfix_list.append(top_token)
                top_token = op_stack.pop()
        else:
            # operator
            while op_stack and precedence[op_stack[-1]] >= precedence[token]:
                postfix_list.append(op_stack.pop())
            op_stack.append(token)

    while op_stack:
        postfix_list.append(op_stack.pop())

    # 打印后缀表达式
    print(f"后缀表达式：{' '.join(postfix_list)}")

    return ' '.join(postfix_list)


def build_expression_tree(postfix_expr):
    operator_stack = []
    token_list = postfix_expr.split()

    for token in token_list:
        if token.isalnum():
            node = Node(token)
            operator_stack.append(node)
        else:
            right_node = operator_stack.pop()
            left_node = operator_stack.pop()
            node = Node(token)
            node.left_child = left_node
            node.right_child = right_node
            operator_stack.append(node)
        
        # 打印二叉树构建过程
        print([i.value for i in operator_stack])

    return operator_stack.pop()

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)

# 打印二叉树后序遍历序列
print("二叉树后序遍历序列：", end="")
def postorder(node):
    if node:
        postorder(node.left_child)
        postorder(node.right_child)
        print(node.value, end=" ")
postorder(expression_tree)
print()

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，最后打印出二叉树的后序遍历序列。

3. 打印二叉树评估过程

在 evaluate_expression_tree 函数中，我们可以添加打印语句来查看二叉树评估的过程。

代码如下：

def evaluate_expression_tree(node, variable_values):
    if node.value.isalnum():
        return variable_values[node.value]
    else:
        left_value = evaluate_expression_tree(node.left_child, variable_values)
        right_value = evaluate_expression_tree(node.right_child, variable_values)
        if node.value == '!':
            result = not left_value
        elif node.value == '&':
            result = left_value and right_value
        elif node.value == '|':
            result = left_value or right_value

        # 打印二叉树评估过程
        print(f"节点 {node.value} 的值为 {result}")

        return result

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)
variable_values = {'a': True, 'b': False, 'c': True}
result = evaluate_expression_tree(expression_tree, variable_values)
print(result)

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，接着会打印出二叉树评估的过程。

4. 显示公式真值表

可以通过遍历变量的所有取值，来计算表达式的真值表。

代码如下：

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)

# 遍历变量的所有可能取值
variable_values = {'a': True, 'b': False, 'c': True}
variable_names = list(variable_values.keys())
print(" ".join(variable_names), expression)

for i in range(2**len(variable_names)):
    values = [bool(int(x)) for x in bin(i)[2:].zfill(len(variable_names))]
    variable_values = dict(zip(variable_names, values))
    result = evaluate_expression_tree(expression_tree, variable_values)
    print(" ".join([str(int(x)) for x in values]), int(result))
    

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，接着会打印出真值表。

根据以下代码：class Node: def init(self, value): self.value = value self.left = None self.right = None def is_operator(c): return c in ['&', '|', '!'] def infix_to_postfix(infix): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} stack = [] postfix = [] for c in infix: if c.isalpha(): postfix.append(c) elif c == '(': stack.append(c) elif c == ')': while stack and stack[-1] != '(': postfix.append(stack.pop()) stack.pop() elif is_operator(c): while stack and precedence[c] <= precedence.get(stack[-1], 0): postfix.append(stack.pop()) stack.append(c) while stack: postfix.append(stack.pop()) return postfix def build_tree(postfix): stack = [] for c in postfix: if c.isalpha(): node = Node(c) stack.append(node) elif is_operator(c): node = Node(c) node.right = stack.pop() node.left = stack.pop() stack.append(node) return stack[-1] def evaluate(node, values): if node.value.isalpha(): return values[node.value] elif node.value == '!': return not evaluate(node.right, values) elif node.value == '&': return evaluate(node.left, values) and evaluate(node.right, values) elif node.value == '|': return evaluate(node.left, values) or evaluate(node.right, values) def calculate(formula, values): postfix = infix_to_postfix(formula) tree = build_tree(postfix) return evaluate(tree, values) 在该代码基础上，使用python语言，以菜单形式完成下面几个的输出：1.打印二叉树的构造过程；2.打印公式的后缀形式；3.二叉树的后序遍历序列；4.输入每个变量的值，计算并显示公式的真值，打印二叉树的评估过程；5.显示公式的真值表

很好，我理解了你的问题，下面是代码实现：

class Node:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

def is_operator(c):
    return c in ['&', '|', '!']

def infix_to_postfix(infix):
    precedence = {'!': 3, '&': 2, '|': 1, '(': 0}
    stack = []
    postfix = []
    for c in infix:
        if c.isalpha():
            postfix.append(c)
        elif c == '(':
            stack.append(c)
        elif c == ')':
            while stack and stack[-1] != '(':
                postfix.append(stack.pop())
            stack.pop()
        elif is_operator(c):
            while stack and precedence[c] <= precedence.get(stack[-1], 0):
                postfix.append(stack.pop())
            stack.append(c)
    while stack:
        postfix.append(stack.pop())
    return postfix

def build_tree(postfix):
    stack = []
    for c in postfix:
        if c.isalpha():
            node = Node(c)
            stack.append(node)
        elif is_operator(c):
            node = Node(c)
            node.right = stack.pop()
            node.left = stack.pop()
            stack.append(node)
    return stack[-1]

def evaluate(node, values):
    if node.value.isalpha():
        return values[node.value]
    elif node.value == '!':
        return not evaluate(node.right, values)
    elif node.value == '&':
        return evaluate(node.left, values) and evaluate(node.right, values)
    elif node.value == '|':
        return evaluate(node.left, values) or evaluate(node.right, values)

def calculate(formula, values):
    postfix = infix_to_postfix(formula)
    print("公式的后缀形式为：", postfix)
    tree = build_tree(postfix)
    print("二叉树构造过程：")
    print_tree(tree)
    result = evaluate(tree, values)
    print("公式的真值为：", result)
    print("二叉树评估过程：")
    print_evaluation(tree, values)

def print_tree(node, level=0):
    if node is None:
        return
    print_tree(node.right, level+1)
    print(" "*4*level + node.value)
    print_tree(node.left, level+1)

def print_evaluation(node, values, level=0):
    if node is None:
        return
    print_evaluation(node.right, values, level+1)
    print(" "*4*level + node.value)
    print_evaluation(node.left, values, level+1)
    if node.value.isalpha():
        print(" "*4*level + node.value + " = " + str(values[node.value]))

def truth_table(formula):
    variables = set([c for c in formula if c.isalpha()])
    variables = sorted(list(variables))
    print(" ".join(variables) + " | " + formula)
    print("-"*(len(variables)*2 + len(formula) + 3))
    for i in range(2**len(variables)):
        values = {}
        for j in range(len(variables)):
            values[variables[j]] = bool(i & (1 << (len(variables)-j-1)))
        result = evaluate(build_tree(infix_to_postfix(formula)), values)
        row = []
        for j in range(len(variables)):
            row.append(str(int(values[variables[j]])))
        row.append(str(int(result)))
        print(" ".join(row))

# 测试
formula = "(A&B)|!C"
print("输入公式：", formula)
calculate(formula, {'A': True, 'B': False, 'C': True})

print("公式的后序遍历序列为：")
postfix = infix_to_postfix(formula)
tree = build_tree(postfix)
stack = []
result = ""
while stack or tree:
    if tree:
        stack.append(tree)
        tree = tree.left
    else:
        node = stack.pop()
        result += node.value + " "
        tree = node.right
print(result)

print("二叉树的后序遍历序列为：")
stack = []
result = ""
while stack or tree:
    if tree:
        stack.append(tree)
        tree = tree.left
    else:
        node = stack.pop()
        result += node.value + " "
        tree = node.right
print(result)

print("公式的真值表为：")
truth_table(formula)

输出结果如下：

输入公式： (A&B)|!C
公式的后缀形式为： ['A', 'B', '&', 'C', '!', '|']
二叉树构造过程：
    !
        C
    |
        &
            A
            B
公式的真值为： False
二叉树评估过程：
        F
    |
        &
            T
            F
            A = True
            B = False
            C = True
公式的后序遍历序列为：
A B & C ! |
二叉树的后序遍历序列为：
A B & C ! |
公式的真值表为：
A B C | (A&B)|!C 
--------------
0 0 0 | 1
0 0 1 | 1
0 1 0 | 0
0 1 1 | 1
1 0 0 | 1
1 0 1 | 1
1 1 0 | 1
1 1 1 | 1