1、如果需要考虑之前的状态就可以使用有限状态机
有限状态机：
Leecode 第8题：
class A:
def __init__(self):
self.sign=1
self.ans=0
self.currentState="empty"
self.dict={
"empty":["empty","sign","digit","end"],
"sign":["end","end","digit","end"],
"digit":["end","end","digit","end"],
"end":["end","end","end","end"]
}
def tocondition(self,s:str,i:int): # 条件转化为数字
if s[i]==' ':
return 0
elif s[i]=='+' or s[i]=='-':
return 1
elif s[i].isdigit():
return 2
else:
return 3
def deal(self,s:str,i:int):
condition = self.tocondition(s,i)
self.currentState=self.dict[self.currentState][condition]
if self.currentState=="sign":
self.sign = -1 if s[i]=='-' else 1
if self.currentState=='digit':
self.ans=self.ans*10+int(s[i])
class Solution:
def myAtoi(self, s: str) -> int:
a = A()
for i in range(len(s)):
a.deal(s,i)
res = a.sign * a.ans
if res<-2**31:
res=-2**31
if res>2**31-1:
res=2**31-1
return res
2、非递归先序遍历
准备一个栈，然后访问根节点，入栈，走到左子树，访问，入栈，左子树没了访问不了，出栈一个元素，走到它的右子树，访问根节点...直到访问不了且栈空了（想用刚刚的元素，就可以想到栈）
stack = []
while root or len(stack)!=0:
if root:
print(root.val)
stack.append(root)
root=root.left
else:
root = stack.pop()
root = root.right
2、层次序列生成二叉树
对当前节点，给他左边加节点，然后入队，右边加节点，然后入队，出队一个元素，对该元素左边加节点...（想到用老之前的元素，就可以想到队列）
def create(self,nums): # 层次建立二叉树
root = Node(nums[0])
q = queue.Queue(len(nums))
q.put(root)
i = 1
while i<=len(nums):
if nums[i]=='null':
i+=1
continue
node = q.get()
node.left = Node(nums[i])
q.put(node.left)
i+=1
if i==len(nums):
break
if nums[i]=='null':
i+=1
continue
node.right = Node(nums[i])
q.put(node.right)
i+=1
return root
def mid(self,root): # 非递归中序遍历
stack = []
while root or len(stack)!=0:
if root:
stack.append(root)
root=root.left
else:
top = stack.pop()
print(top.val)
root = top.right
def pre(self,root): # 非递归先序遍历
stack = []
while root or len(stack)!=0:
if root:
print(root.val)
stack.append(root)
root=root.left
else:
root = stack.pop()
root = root.right
3
标签：digit,end,self,sign,力扣,return,root From： https://www.cnblogs.com/pjishu/p/16739869.html