Convert Infix to Postfix using Stack in Python

# Define the Stack class
class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        if not self.isEmpty():
            return self.items.pop()
        else:
            return None

    def peek(self):
        if not self.isEmpty():
            return self.items[-1]
        else:
            return None

    def isEmpty(self):
        return len(self.items) == 0

# This function returns the relative precedence of the operators we have
# e.g divide (/) has higher precedence than multiply (*)
def precedence(char):
    if char == '^':
        return 3
    elif char == '/' or char == '*':
        return 2
    elif char == '+' or char == '-':
        return 1
    else:
        return 0

# Function check whether the given character is Operand or not
def isOperand(char):
    return (ord(char) >= ord('a') and ord(char) <= ord('z')) or (ord(char) >= ord('A') and ord(char) <= ord('Z')) or (ord(char) >= ord('0') and ord(char) <= ord('9'))

# Function for converting an infix expression to postfix expression
def infixToPostfix(s, st): # s is the infix expression and st is the stack
    postFix = []
    for i in range(len(s)):
        if isOperand(s[i]):
            postFix.append(s[i])
        elif s[i] == '(':
            st.push(s[i])
        elif s[i] == ')':
            while st.peek() != '(':
                postFix.append(st.peek())
                st.pop()
            st.pop()  # Pop the '(' from the stack
        else:
            while (not st.isEmpty()) and (precedence(s[i]) <= precedence(st.peek())):
                postFix.append(st.pop())
            st.push(s[i])

    while not st.isEmpty():
        postFix.append(st.peek())
        st.pop()

    return ' '.join(postFix)

# Example usage:
infix_expression = "a + b * (c - d) / e"
stack = Stack()
postfix_expression = infixToPostfix(infix_expression, stack)
print(postfix_expression)