leetcode之Serialize and Deserialize Binary Tree

来源：互联网发布：数组大小可以任意改变编辑：程序博客网时间：2024/05/01 07:09

这题主要还是运用了广度搜索优先的方法。一层一层的遍历来获得每层的数据。然后依次安装上去。代码如下：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Codec:    def serialize(self, root):        """Encodes a tree to a single string.                :type root: TreeNode        :rtype: str        """        if not root:            return ''        stack1 = [root]        list1 = [root.val]        while len(stack1) != 0:            stack2 = []            for i in stack1:                if i:                    stack2.append(i.left)                    stack2.append(i.right)            for i in stack2:                if i:                    list1.append(i.val)                else:                    list1.append(None)            stack1 = stack2            if stack2 == []:                return str(list1)            def deserialize(self, data):        """Decodes your encoded data to tree.                :type data: str        :rtype: TreeNode        """        if data == '':            return None        data = data[1:-1]        data = data.split(', ')        root = TreeNode(int(data[0]))        data = data[1:][::-1]        stack1 = [root]        while data != []:            stack2 = []            for i in stack1:                if data[-1] == 'None':                    data.pop()                    i.left = None                    if data[-1] == 'None':                        data.pop()                        i.right = None                    else:                        i.right = TreeNode(int(data.pop()))                        stack2.append(i.right)                else:                    i.left = TreeNode(int(data.pop()))                    stack2.append(i.left)                    if data[-1] == 'None':                        i.right = None                        data.pop()                    else:                        i.right = TreeNode(int(data.pop()))                        stack2.append(i.right)            stack1 = stack2        return root

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