331: Time 98%, Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nan explicit stack, and instead just uses a single variable degree to keep
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nan explicit stack, and instead just uses a single variable degree to keep track of the outDegree (children) - inDegree (parent). It also removes the need for the stack and nodes lists used in the previous solution, which results in less memory usage and faster execution time.\n\nThe time complexity of this solution is O(n), where n is the number of nodes in the binary tree. The space complexity is O(1), as we are only using a single variable to keep track of the degree.\n\n# Complexity\n- Time complexity:\n98%\n\n- Space complexity:\n75.32%\n\n# Code\n```\nclass Solution:\n    def isValidSerialization(self, preorder: str) -> bool:\n        # Initialize the outDegree (children) - inDegree (parent) to 1\n        degree = 1\n        \n        # Iterate through the nodes in the preorder traversal\n        for node in preorder.split(\',\'):\n            degree -= 1 # Decrement the degree by 1 for each node\n            \n            if degree < 0: # If the degree is negative, return False\n                return False\n            \n            if node != \'#\': # If the node is not a leaf node\n                degree += 2 # Increment the degree by 2 for each non-leaf node\n            \n        # If the final degree is 0, the tree is valid, else invalid\n        return degree == 0\n\n```
['String', 'Stack', 'Tree', 'Python', 'Python3']