[LeetCode] 331. Verify Preorder Serialization of a Binary Tree_Medium tag: stack

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:

Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: "1,#"
Output: false

Example 3:

Input: "9,#,#,1"
Output: false

这个题目的思路参考Solution, 本质上就是看这个tree 是否valid, 然后就要满足可以有的children数>= 0 , 最后== 0, 那么最开始slot = 1, 然后for loop, 每
多一个数字, 表明可以多一个child, 如果是#表明少一个child, 最后children数等于0 即可.

就是用Stack,每次碰到#, 去看stack[-1]是否为#, 如果是的话表明stack[-2]的children满了, 所以将stack.pop()执行两次, 并且循环, 然后把#放到stack里面, 这里在stack.pop()两次
的中间, 加入一个stack 非空的判断, 是针对"##" "###"这样的情况来的.最后len(stack) == 1 and stack[-1] == '#'

Code: T: O(n)   S: O(n)

 class Solution:
     def preorderValid(self, preorder):
         stack, preorder = [], preorder.split(',')
         for p in preorder:
            while (p == '#' and stack and stack[-1] == '#'):
                  stack.pop()
                  if not stack: return False
                  stack.pop()
            stack.append(p)
         return len(stack) == 1 and stack[-1] == '#'

Code: T: O(n)   S: O(1)

class Solution:
    def preorderValid(self, preorder):
        slot, preorder = 1, preorder.split(',')
        for p in preorder:
            if slot == 0: return False # means that should have no children anymore
            if p == '#':
                slot -= 1
            else:
                slot += 1
        return slot == 0