C++ Solution | Easy to understand code with explanation | O(n) Time |  O(1) Space
The idea here is to compute the required nodes for a given node.\nFor 1 non-null node -> required node count = 2 (including null) .\nFor 1 null node -> no more 
The idea here is to compute the required nodes for a given node.\nFor 1 non-null node -> required node count = 2 (including null) .\nFor 1 null node -> no more node is required as it is closing that branch.\nOur aim to check if we have sufficient node count available at every node.\n\nWe start iterating, and ignore all the comma character.\nWhenever we get a non- null node, we reduce reqirement by 1 and also add +2 because this node will further need 2 more nodes(null/not null) to close\nIf we get a null node, we simply reduce requirement by one.\nAlso check if requirement is always non- negative\nand after complete check, requirements should be 0\n\n```\nbool isValidSerialization(string preorder) {\n        int n = preorder.size();\n        int req =1;\n        for(int i=0;i<n;i++){\n            if(preorder[i]==\',\') continue;\n            else if(preorder[i]==\'#\'){\n                req--;\n                if(req<0)  return 0;\n            }\n            else{\n                while(i<n-1 && preorder[i+1]!=\',\')i++;\n                req--;\n                if(req<0) return 0;\n                req+=2;\n            }\n        }\n        return req==0 ;\n    }
['C', 'C++']