ParseTree.py

'''
构造一棵解析树
需要调用之前写过的Stack文件和BinaryTree文件
'''

from Stack import Stack
from BinaryTree import BinaryTree
import operator

# 构造解析树
def buildParseTree(fpexp):
    fplist = fpexp.split()
    pStack = Stack()
    eTree = BinaryTree('')
    pStack.push(eTree)
    currentTree = eTree
    for i in fplist:
        if i == '(':
            currentTree.insertLeft('')
            pStack.push(currentTree)
            currentTree = currentTree.getLeftChild()
        elif i not in ['+', '-', '*', '/', ')']:
            currentTree.setRootVal(int(i))
            parent = pStack.pop()
            currentTree = parent
        elif i in ['+', '-', '*', '/']:
            currentTree.setRootVal(i)
            currentTree.insertRight('')
            pStack.push(currentTree)
            currentTree = currentTree.getRightChild()
        elif i == ')':
            currentTree = pStack.pop()
        else:
            raise ValueError
    return eTree

# 递归实现两个叶结点的运算
def evaluate(parseTree):
    opers = {'+':operator.add, '-':operator.sub, '*':operator.mul, '/':operator.truediv}

    leftC = parseTree.getLeftChild()
    rightC = parseTree.getRightChild()

    if leftC and rightC:
        fn = opers[parseTree.getRootVal()]
        return fn(evaluate(leftC), evaluate(rightC))
    else:
        return parseTree.getRootVal()

# 递归实现树的后序遍历
def postorder(tree):
    if tree != None:
        postorder(tree.getLeftChild())
        postorder(tree.getRightChild())
        print(tree.getRootVal())

# 利用后序遍历实现两个叶结点的运算
def postordereval(tree):
    opers = {'+': operator.add, '-': operator.sub, '*': operator.mul, '/': operator.truediv}
    res1 = None
    res2 = None
    if tree:
        res1 = postordereval(tree.getLeftChild())
        res2 = postordereval(tree.getRightChild())
        if res1 and res2:
            return opers[tree.getRootVal()](res1, res2)
        else:
            return tree.getRootVal()

# 递归实现中序遍历
def inorder(tree):
    if tree != None:
        inorder(tree.getLeftChild())
        print(tree.getRootVal())
        inorder(tree.getRightChild())

# 因为中序遍历会丢失括号信息, 因此尝试构造一个函数回复解析表达式
def printexp(tree):
    sVal = ''
    if tree:
        sVal = '(' + printexp(tree.getLeftChild())
        sVal = sVal + str(tree.getRootVal())
        sVal = sVal + printexp(tree.getRightChild()) + ')'
    return sVal


pt = buildParseTree("( ( 10 + 5 ) * 3 )")
# pt.preorder()
# postorder(pt)
# ans = evaluate(pt)
# print(ans)
# print(postordereval(pt))
# inorder(pt)
# print(pt)
sVal = printexp(pt)
print(sVal)