Convert the input infix expression to postfix expression. Calculate the expression according to the postfix expression.

Sure, I can help you with that. For the first part, to convert an infix expression to a postfix expression, we can use the Shunting Yard Algorithm. For the second part, we can use a stack to evaluate the postfix expression. Do you need more information on how to implement this algorithm?

相关问题

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，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

1. 打印二叉树的构造过程：

假设给定的公式为：(A&B)|(!C&D)

构造二叉树的过程如下：

|
/ \
& D
/ \
A !
/
C

2. 打印公式的后缀形式：

给定的公式为：(A&B)|(!C&D)

后缀形式为：AB&CD&!|

3. 二叉树的后序遍历序列：

假设给定的公式为：(A&B)|(!C&D)

二叉树的后序遍历序列为：A B & C ! D & |

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：

假设给定的公式为：(A&B)|(!C&D)，输入变量的值为 A=True, B=False, C=True, D=True

二叉树的评估过程如下：

|
/ \
& &
/ \ / \
A B ! D
|
C

公式的真值为：True

5. 显示公式的真值表：

给定的公式为：(A&B)|(!C&D)

| A | B | C | D | (A&B)|(!C&D) |
| --- | --- | --- | --- | ------------ |
| T | T | T | T | T |
| T | T | T | F | T |
| T | T | F | T | T |
| T | T | F | F | T |
| T | F | T | T | F |
| T | F | T | F | F |
| T | F | F | T | F |
| T | F | F | F | F |
| F | T | T | T | F |
| F | T | T | F | F |
| F | T | F | T | F |
| F | T | F | F | F |
| F | F | T | T | F |
| F | F | T | F | T |
| F | F | F | T | F |
| F | F | F | F | F |

这段代码有什么问题class Node: def __init__(self, value): self.value = value self.left_child = None self.right_child = None def build_tree(postfix_expr): opstack = [] token_list = postfix_expr.split() for token in token_list: if token in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" : opstack.append(Node(token)) else: right_child = opstack.pop() left_child = opstack.pop() new_node = Node(token) new_node.left_child = left_child new_node.right_child = right_child opstack.append(new_node) return opstack.pop() def evaluate_tree(tree, var_dict): if tree.left_child and tree.right_child: left_value = evaluate_tree(tree.left_child, var_dict) right_value = evaluate_tree(tree.right_child, var_dict) return evaluate_op(tree.value, left_value, right_value) else: return var_dict[tree.value] def evaluate_op(op, left_value, right_value): if op == "&": return left_value and right_value elif op == "|": return left_value or right_value else: return not left_value def calculate(expr, var_dict): postfix_expr = infix_to_postfix(expr) tree = build_tree(postfix_expr) result = evaluate_tree(tree, var_dict) return result

这段代码的问题在于没有提供 infix_to_postfix 函数的实现。因此，我们无法确定整个程序的正确性。另外，我们也无法确定 var_dict 字典包含的键和值，因为该字典在代码中没有被定义。需要注意的是，如果 var_dict 字典中不存在某个变量，则 evaluate_tree 函数会引发 KeyError 异常。