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## Problem

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


## Code

class Solution {
public boolean isValidSerialization(String preorder) {
// 每个非叶子节点都有2出度和1入度, 除了根节点
// 每个叶子节点都有0出度和1入度
String[] nodes = preorder.split(",");
// 让根节点有一个入度
// degree = outDegree - inDegree
int degree = 1;
for(String node : nodes){
// 新来一个节点，要消耗一个度
// 如果此时没有度可以用，那么就返回false
if(--degree < 0){
return false;
}
// 如果不是叶子节点，那就要增加两个出度
if(!node.equals("#")){
degree += 2;
}
}
return degree == 0;
}
}

class Solution {
public boolean isValidSerialization(String preorder) {
Stack<String> stack = new Stack<>();

for (String s : preorder.split(",")){
while (!stack.isEmpty() && stack.peek().equals("#") && s.equals("#")) {
stack.pop();
if(stack.isEmpty()){
return false;
}
stack.pop();
}
stack.push(s);
}

return stack.size() == 1 && stack.peek().equals("#");
}
}