leetcode 之 Verify Preorder Serialization of a Binary Tree

题目描述：

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   
   3     2
  /    / 
 4   1  #  6
/  /    / 
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

即 给定一颗二叉树的前序遍历串，判断这个串是不是合法的。

分析：1. 前序遍历的即为 根左右遍历，首先访问根节点，然后依次左子树和右子树。

　　　2. 对于9,3,4,#,#,1,#,#,2,#,6,#,# 这个串来说，6, #,# 一定是一个叶子节点，将其移除后，替换为"#"

　　　3. 这时串变为： 9,3,4,#,#,1,#,#,2,#,#，继续第二步操作

使用堆栈保存输入的串，当堆栈后三个元素为 x## 时， 移除这三个元素，并加入#， 最后如果堆栈长度为1 且 值为# ，则返回true

代码如下：

class Solution(object):
    def isValidSerialization(self, preorder):
        """
        :type preorder: str
        :rtype: bool
        """
        l = len(preorder)
        if l == 1 and preorder[0] == "#":
            return True
        if preorder[0] == "#":
            return False
        ls = preorder.split(",")
        stack = []
        for i in ls:
            stack.append(i)
            while len(stack) >= 3 and stack[-2] == "#" and stack[-1] == "#" and stack[-3] != "#":
                stack.pop()
                stack.pop()
                stack.pop()
                stack.append("#")
                
        return len(stack) == 1 and stack[0] == "#"