题目大意：验证二叉树前序遍历的序列化是否合法。

https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/description/

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 nullpointer.

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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Solution: Recursion

1. If a node is not null, it must has two children, thus verify left subtree and right subtree recursively.
2. If a not is null, the current char must be ‘#’

Time complexity: O(n)

Space complexity: O(h)

C++

// Author: Huahua
// Running time: 5 ms
class Solution {
public:
  bool isValidSerialization(string preorder) {
    int pos = 0;    
    return isValid(preorder, pos) && pos == preorder.length();
  }
private:
  bool isValid(const string& s, int& pos) {    
    if (pos >= s.length()) return false;
    if (isdigit(s[pos])) {      
      while (isdigit(s[pos])) ++pos;
      return isValid(s, ++pos) && isValid(s, ++pos);
    }
    return s[pos++] == '#';
  }
};

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