【LeetCode】331. Verify 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

转载discuss上一个思路，连续遇到两个“#”，说明这是叶子节点，删去他们及他们的根节点，用一个’#‘代替，直到最后剩一个’#‘说明是二叉树。否则，返回false。

class Solution {public:    bool isValidSerialization(string preorder) {        int len = preorder.length();        stack<char> stk;        int count =0;        for(int i=0; i < len;i++)        {            if(isdigit(preorder[i]))            {                 while(i<len&&isdigit(preorder[i]))                  {                      i++;                  }                  i--;                stk.push(preorder[i]);            }            else if(preorder[i]=='#')            {                stk.push(preorder[i]);                while(stk.top()=='#'&&stk.size()>=2)                {                    stk.pop();                    if(stk.top()=='#')                    {                        stk.pop();                        if(stk.empty())return false;                          if(stk.top()=='#')return false;                         stk.pop();                        stk.push('#');                    }else{                        stk.push('#');                        break;                    }                }            }        }        if(stk.top()=='#'&&stk.size()==1)          return true;         else return false;           }};