#构建二叉树 def postfix_binary_tree(postfix,data): stack = Stack() for token in postfix: if token not in {'∧', '∨', '¬'}: cur= TreeNode(token) cur.print_tree(cur) cur_result = cur.evaluate_formula(cur,data) print("当前的真值为：",cur_result) stack.push(cur) else: left_node = None right_node = None if stack: right_node = stack.pop() if token != '¬': if stack: left_node = stack.pop() cur1 = TreeNode(token) cur1.left = left_node cur1.right = right_node cur1.print_tree(cur1) cur1_result=cur1.evaluate_formula(cur1,data) print("当前的真值为：",cur1_result) stack.push(cur1) tree = stack.pop() return tree

这段代码实现了根据后缀表达式构建二叉树的功能。具体实现过程如下：

1. 定义一个栈，用于存储节点。

2. 遍历后缀表达式中的每个元素。

3. 如果当前元素不是运算符，即为操作数，创建一个节点，并将其压入栈中。

4. 如果当前元素是运算符，弹出栈顶的两个节点作为左右子节点，创建一个新节点，并将其作为根节点，左右子节点分别指向弹出的两个节点。将新节点压入栈中。

5. 重复步骤2-4，直到后缀表达式的所有元素都被扫描完毕。

6. 最后栈中仅剩一个节点，即为构建的二叉树的根节点。

其中，TreeNode类表示二叉树节点，包括节点值、左右子节点和节点的打印和求值方法；Stack类表示栈，包括压栈、弹栈和判断栈是否为空的方法。

需要注意的是，在代码中调用了节点的 evaluate_formula 方法，用于计算节点的真值。该方法需要传入一个字典 data，表示变量的取值情况。例如，如果变量 x 的取值为 True，则 data 字典中应该包含键值对 {'x': True}。

另外，代码中的运算符集合 {'∧', '∨', '¬'} 表示逻辑运算符 and、or、not。

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。

class TreeNode: def __init__(self, val=None, left=None, right=None): self.val = val self.left = left self.right = right def infix_to_postfix(infix): operators = {'(': 0, ')': 0, 'NOT': 1, 'AND': 2, 'OR': 3} stack = [] postfix = [] for token in infix: if token in operators: if token == '(': stack.append(token) elif token == ')': while stack[-1] != '(': postfix.append(stack.pop()) stack.pop() else: while stack and operators[stack[-1]] >= operators[token]: postfix.append(stack.pop()) stack.append(token) else: postfix.append(token) while stack: postfix.append(stack.pop()) return postfix def postfix_to_tree(postfix): stack = [] for token in postfix: if token in {'NOT', 'AND', 'OR'}: right = stack.pop() if token == 'NOT': stack.append(TreeNode('NOT', None, right)) else: left = stack.pop() stack.append(TreeNode(token, left, right)) else: stack.append(TreeNode(token)) return stack.pop() def evaluate(root, values): if root.val in values: return values[root.val] elif root.val == 'NOT': return not evaluate(root.right, values) elif root.val == 'AND': return evaluate(root.left, values) and evaluate(root.right, values) elif root.val == 'OR': return evaluate(root.left, values) or evaluate(root.right, values) def print_tree(root, level=0): if root: print_tree(root.right, level + 1) print(' ' * 4 * level + '->', root.val) print_tree(root.left, level + 1) infix = input('请输入命题演算公式：').split() postfix = infix_to_postfix(infix) root = postfix_to_tree(postfix) print('后缀表达式：', postfix) print('二叉树构造过程：') print_tree(root) print('真值表：') variables = list(set(filter(lambda x: x not in {'NOT', 'AND', 'OR'}, infix))) for values in itertools.product([True, False], repeat=len(variables)): values = dict(zip(variables, values)) result = evaluate(root, values) print(values, '->', result)其中有错误NameError: name 'itertools' is not defined。请修改

在代码的开头处添加`import itertools`即可解决该错误：

import itertools

class TreeNode:
    def __init__(self, val=None, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def infix_to_postfix(infix):
    operators = {'(': 0, ')': 0, 'NOT': 1, 'AND': 2, 'OR': 3}
    stack = []
    postfix = []
    for token in infix:
        if token in operators:
            if token == '(':
                stack.append(token)
            elif token == ')':
                while stack[-1] != '(':
                    postfix.append(stack.pop())
                stack.pop()
            else:
                while stack and operators[stack[-1]] >= operators[token]:
                    postfix.append(stack.pop())
                stack.append(token)
        else:
            postfix.append(token)
    while stack:
        postfix.append(stack.pop())
    return postfix

def postfix_to_tree(postfix):
    stack = []
    for token in postfix:
        if token in {'NOT', 'AND', 'OR'}:
            right = stack.pop()
            if token == 'NOT':
                stack.append(TreeNode('NOT', None, right))
            else:
                left = stack.pop()
                stack.append(TreeNode(token, left, right))
        else:
            stack.append(TreeNode(token))
    return stack.pop()

def evaluate(root, values):
    if root.val in values:
        return values[root.val]
    elif root.val == 'NOT':
        return not evaluate(root.right, values)
    elif root.val == 'AND':
        return evaluate(root.left, values) and evaluate(root.right, values)
    elif root.val == 'OR':
        return evaluate(root.left, values) or evaluate(root.right, values)

def print_tree(root, level=0):
    if root:
        print_tree(root.right, level + 1)
        print(' ' * 4 * level + '->', root.val)
        print_tree(root.left, level + 1)

infix = input('请输入命题演算公式：').split()
postfix = infix_to_postfix(infix)
root = postfix_to_tree(postfix)
print('后缀表达式：', postfix)
print('二叉树构造过程：')
print_tree(root)
print('真值表：')
variables = list(set(filter(lambda x: x not in {'NOT', 'AND', 'OR'}, infix)))
for values in itertools.product([True, False], repeat=len(variables)):
    values = dict(zip(variables, values))
    result = evaluate(root, values)
    print(values, '->', result)