序列化二叉树

#python非递归解法，可能思路比较麻烦
class Solution:
    def Serialize(self, root):
        # write code here
        #先序遍历二叉树
        if not root:
            return ['#']
        stack=[]
        node=root
        value=[]
        while node&nbs***bsp;stack:
            while node:
                value.append(str(node.val)+'!')
                stack.append(node)
                node=node.left
            value.append('#')
            node=stack.pop()
            node=node.right
        value.append('#')
        return ''.join(value)
    
    def Deserialize(self, s):
        # write code here
        if s=='#':
            return None
        node1=TreeNode(0)
        root=node1
        stack=[]
        i=0
        temp=0
        while len(s)>1&nbs***bsp;stack:
            if s[i]=='!':
                val=int(s[:i])
                s=s[i+1:]
                node=TreeNode(val)
                stack.append(node)
                if temp==0:
                    node1.left=node
                    node1=node
                else:
                    node1.right=node
                    node1=node
                i=0
                temp=0
            elif s[i]=='#':
                s=s[i+1:]
                node1=stack.pop(-1)
                temp=1
                i=0
            else:
                i=i+1
        return root.left