EASY C++ SOLTUION USING STACK
\n# Approach\nthe first character of the preorder string is a comma, return false because a valid preorder serialization should not start with a comma.\nIf the 
\n# Approach\nthe first character of the preorder string is a comma, return false because a valid preorder serialization should not start with a comma.\nIf the preorder string consists of a single character which is \'#\', return true because it represents a valid serialization of an empty tree.\nInitialize a stack to simulate the construction of the binary tree nodes during the traversal.\n\nIterate over the characters in the preorder string:\n\nIf a comma is encountered, it indicates the end of a node\'s value. Push the current temp string representing the node\'s value onto the stack.\nIf a character other than a comma is encountered, append it to the temp string.\nAfter processing each node, check if there are more than two consecutive \'#\' symbols on the stack. If so, it means there are two leaf nodes without a parent, which is not a valid serialization. Return false in this case.\n\nIf there are three consecutive \'#\' symbols on the stack, replace them with a single \'#\' symbol to simulate collapsing them into a single leaf node.\n\nRepeat steps 4 and 5 until the stack has at most one element.\n\nFinally, check if there is exactly one element left on the stack, and if that element is \'#\'. If both conditions are true, return true; otherwise, return false.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n\n    bool isValidSerialization(string preorder) {\n        if(preorder[0]==\',\') return false;\n        if(preorder[0]==\'#\') \n        {\n            if(preorder.length()==1)\n            {\n                return true;\n            }    \n            return false;\n        }\n        preorder.push_back(\',\');\n        stack<string>st;\n        string temp="";\n        for(int i = 0;i<preorder.length();i++)\n        {\n            if(preorder[i]==\',\'){\n            st.push(temp);\n            while(st.size()>2)\n            {\n                string s1=st.top();\n                st.pop();\n                string s2=st.top();\n                st.pop();\n                string s3=st.top();\n                st.pop();\n                if((s1=="#" && s2=="#" && s3=="#"))\n                {\n                    return false;\n                }\n                if((s1=="#" && s2=="#" && s3!="#"))\n                {\n                    st.push("#");\n                }\n                else\n                {\n                    st.push(s3);\n                    st.push(s2);\n                    st.push(s1);\n                    break;\n                }\n            }\n            temp.clear();\n        }\n        else\n        {\n            temp.push_back(preorder[i]);\n        }\n    }\n    return st.size()==1 and st.top()=="#";\n    }\n};\n```
['String', 'Stack', 'Binary Tree', 'C++']