331. Verify Preorder Serialization of a Binary Tree ---------M ~~~~

 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

A:

思路很清晰， 就是保持一个  tree depth的stack，来记录我们是在走该节点的左子树，还是右子树

class Solution {
public:
    bool isValidSerialization(string preorder) {
        stringstream check(preorder);
        vector<string> tokens;
        string tmp;
        while(getline(check, tmp, ',')){
            tokens.push_back(tmp);
        }
        if(tokens.size()==1 && tokens.back()=="#") // if just an empty tree
            return true;
        
        vector<char> V;// only 'L' for going left, 'R' for right
        for(int i = 0; i< tokens.size(); i++){
            string token = tokens[i];
            if(i>0 && V.empty()) // if not first time, but it is empty
                return false;
            
            if(token == "#"){
                if(V.empty())
                    return false;
                while(V.size()>0 && V.back()=='R'){
                    V.pop_back();
                }
                if(V.size() > 0 && V.back()=='L'){
                    V[V.size()-1] = 'R';
                }
            }else{
                V.push_back('L');
            }
        }
        return V.empty();
    }
};

---------Errors:

没有考虑 “#” 的情况。其实也是可以的

这个只是用了count 

理论依据是： tree里，node 的数量 一直  >=   # , 然后最后小于一个。

class Solution {
public:
    bool isValidSerialization(string preorder) {
        int count = 0;
        for(int i = 0 ; i < preorder.size() ; i++){
            if(preorder[i] == ',') 
                continue;
            
            if(count < 0) 
                return false;
            
            if(preorder[i] == '#')
                count--;
            else{
                while(i < preorder.size() && preorder[i]!=',')
                    i++;
                count++;
            }
        }
        return count == -1;
    }
};