leetcode刷题，总结，记录，备忘331

leetcode331Verify Preorder Serialization of a Binary Tree

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

这道题目也是自己没相处方法，去翻了别人的博客。二叉树的特性，就是每在一个空节点上插入1个叶节点的话，原先的空节点消失，新增2个空节点，所以二叉树的空节点最终比所有有值的节点多1，由这点入手。统计字符串序列中空节点和有值节点的个数，在不考虑最后一个节点的情况，空节点个数和有值节点个数应该相同，并且最后一个节点一定为#。除去只有一个节点并且该节点是#的情况，其他情况下，第一个节点必定不为#。

class Solution {
public:
    bool isValidSerialization(string preorder) {
        if (preorder.empty()) 
        {
           return false; 
        }
        
        istringstream input(preorder);
        vector<string> vs;
        string temp;
        int count = 0;
        
        while (getline(input, temp, ','))
        {
            vs.push_back(temp);
        }
        
        for (int i = 0; i < vs.size() - 1; ++i) 
        {
            if (vs[i] == "#") 
            {
                if (count == 0) 
                {
                    return false;
                }
                else 
                {
                    --count;
                }
            }
            else 
            {
               count++; 
            }
        }
        
        
        if (count == 0 && vs.back() == "#")
        {
            return true;
        }
        else
        {
            return false;
        }
    }
};