Serialize and deserialize binary tree code#

Return left + ',' + right + ',' + to_string(root - > val) ĭeserialization- Now to deserialize we again use postorder. String right = serialize(root - > right) Now return the concatenation of left, right, and root values separated by “,” into one string. If root is null return the special character “#”.ĭeclare two variables, left and right to store the serialized string from the left and right subtree respectively. Serialization- In this approach, we will first serialize the left subtree, then the right subtree, and then the root, all recursively. While serializing, in the case of encountering a null value we will use a special character like “#” instead and for the value, we convert them into a string. We will be using different binary tree traversal methods to solve this problem. Postorder serialized string: “#,#,#,35,#,#,6,29,18,#,#,#,15,11,4”ĭeserializing the tree using one of the two above strings we should get the same tree. We can use various algorithms to do the serialization and deserialization to convert a binary tree into a string and then obtain the binary tree from the string.

All the elements in three strings are splited using ','.Īnswer (1) array (22) backtracking (1) BFS (15) binary search (12) binary tree (7) bit manipulations (9) BST (2) bubble sort (1) bucket sort (1) c++ (1) collision (1) common prefix (1) compare (1) design (1) DFS (18) dict (2) divide and conquer (2) DP (13) dynamic programming (1) graph (2) hash (7) hash function (1) hash table (7) heap (3) heap sort (3) Index (2) inorder (2) insert sort.In this problem we are to convert a data structure or object into a sequence of bits, by means of which it can be stored in a file or memory buffer, or transmitted across a network (this is called serialization), which then can be reconstructed into the same using deserialization. The 'vaule string' is the values of each node (using preorder order) in string format. The 'inorder string' is the indices of each node in inorder order. The 'preorder string' is the indices of each node in preorder order, actually, I have given the sequence from 1 to the number of nodes in this string for simplicity. So, our encoding format is: 'preorder string inorder string value string'. Note that in this problem, there might be duplicate values in different nodes, we have to use the indices of each node for the traversal. Please take a look at my previous post for details. Therefore, we could use preorder and inorder traversal to reconstruct the binary tree. Particularly, we have a binary tree, rather than the level by level traversal, we still have three depth-first traversal: preorder, inorder, and postorder. Although it is not a quite efficient method, it still provides good practice of binary tree and binary tree traversal. Secondly, I'l like to show the implementation using a different way. For the deserialization, we could still keep two queues for the BFS, and keep track of the node we are expanding. Here in my solution, I choose to use two queues, to store nodes in current level, and nodes in next level, respectively. Since we are 'searching' the tree level by level, DFS is usually a good way to do so. We will encode the node value level by level, and only encode the 'Null' (or 'None' in python) node when it is a leaf node. Let's first see the simple way of doing this.

Serialize and deserialize binary tree code#

If you would like to see the simple version which is the same way of LeetCode OJ, please check out the pyhton code below.

In this problem, I have implemented a differrent way of serialization (in C++ code shown below).