# # 【LeetCode】331. Verify Preorder Serialization of a Binary Tree 解题报告（Python）

## # 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
``````

## # 解题方法

``````如：”9,3,4,#,#,1,#,#,2,#,6,#,#” 遇到x # #的时候，就把它变为 #

9,3,4,#,# => 9,3,# 继续读
9,3,#,1,#,# => 9,3,#,# => 9,# 继续读
9,#2,#,6,#,# => 9,#,2,#,# => 9,#,# => #
``````
``````class Solution(object):
def isValidSerialization(self, preorder):
"""
:type preorder: str
:rtype: bool
"""
stack = []
for node in preorder.split(','):
stack.append(node)
while len(stack) >= 3 and stack[-1] == stack[-2] == '#' and stack[-3] != '#':
stack.pop(), stack.pop(), stack.pop()
stack.append('#')
return len(stack) == 1 and stack.pop() == '#'
``````

1. 所有的非空节点提供2个出度和1个入度（根除外）
2. 所有的空节点但提供0个出度和1个入度

``````class Solution(object):
def isValidSerialization(self, preorder):
"""
:type preorder: str
:rtype: bool
"""
nodes = preorder.split(',')
diff = 1
for node in nodes:
diff -= 1
if diff < 0:
return False
if node != '#':
diff += 2
return diff == 0
``````

2018 年 3 月 13 日