python 二叉树实现带括号的四则运算

#!/usr/bin/python
#* encoding=utf-8
s = "20-5*(0+1)*5^(6-2^2)" c = 0
top = [0,s[c],0]
op = [["0","1","2","3","4","5","6","7","8","9"],["+","-"],["*","/"],["^"]] def getLev(ch):
for c1 in range(0, len(op)):
for c2 in range(0, len(op[c1])):
if (op[c1][c2]==ch):
return c1
elif (len(ch)>1):
match = 0
for c3 in range(0, len(ch)):
if (getLev(ch[c3])>=0):
match+=1
if (match==len(ch)):return c1
return -1 def makeTree(root):
global c
global s c += 1
if (c>=len(s)):
return root if (s[c]=="("):
c+=1
node = [0, s[c], 0]
node = makeTree(node)
elif (s[c]==")"):
return root
else: node=[0, s[c], 0] levRoot = getLev(root[1])
levCur = getLev(node[1])
print levRoot, levCur, root[1], node[1] if (levCur>=levRoot):
if ((levRoot==0 and levCur!=levRoot)
or (levRoot!=0 and levCur==levRoot)):
node[0] = root
root = node
return makeTree(root)
elif (levRoot==0 and levCur==0):
root[1] += node[1]
return makeTree(root)
else:
node[0] = root[2]
root[2] = makeTree(node)
return makeTree(root)
else:
if (levCur==0 or node[0]!=0):
root[2] = node
return makeTree(root)
else:
c-=1
return root top = makeTree(top)
#print top def getTree(node):
ret = [] if (node[0]!=0):
_tmp = getTree(node[0])
for c in range(0, len(_tmp)):
ret.append(_tmp[c]) if (node[2]!=0):
_tmp = getTree(node[2])
for c in range(0, len(_tmp)):
ret.append(_tmp[c]) ret.append(node[1])
return ret exp = getTree(top)
print exp def calc():
stack=[] for c in range(0, len(exp)):
if (exp[c]>='0' and exp[c]<='9'):
stack.append(exp[c])
else:
op = exp[c]
n2 = stack.pop()
n1 = stack.pop()
if (op!="^"):
v = n1+op+n2
else:
v = "pow(%s,%s)"%(n1,n2)
print v, eval(v)
stack.append("%s"%eval(v)) return stack.pop() print calc()

  

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