Hello everybody, [INTRO] this blog series is the first of many as my attempt to become an expert on code-forces before my fourth year of university is over. I'm currently studying in my third year and about to finish my third year of studying computer engineering. i know I might be a bit late to the party but better late than never!

this is proof to myself that I can di anything. [TOPIC] Serialization of a binary tree: the problem I tried to solve using bfs, having to cover each level where a level is at most 2^n nodes, making the serialization process intensive to say the least Actually It is O(l * 2^l)

def serialize(self, root):
        """Encodes a tree to a single string.
        
        :type root: TreeNode
        :rtype: str
        """
        def get_height(node):
            if not node:
                return 0
            return max(get_height(node.left), get_height(node.right)) +1
        height = get_height(root) 
        res = []
        q = deque()
        emp = "1001"
        if not root:
            res.append(emp)
            return "".join(res)
        q.appendleft(root)	
        for l in range(height + 1):
            len_q = len(q)
            for x in range(int(pow(2, l))):
                curr =  q.pop()
                res.append(str(curr.val))
                if  curr.left:
                    q.appendleft(curr.left)
                else:
                    q.appendleft(TreeNode(1001))
                if curr.right:
                    q.appendleft(curr.right)
                else:
                    q.appendleft(TreeNode(1001))
                res.append("_")
        res = "".join(res)
        return res
        
                

I learned it's just like constructing a tree from pre/post-order. but I don't remember how it's done so I'll study it and return with another post!