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 studying in my third year and about to finish my third year 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 do 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!