Last Updated: 1 Dec, 2020

Serialize and Deserialize Binary Tree

Hard

Asked in companies

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Problem statement

You have been given a binary tree of integers. You are supposed to serialize and deserialize (refer to notes) the given binary tree.

You can choose any algorithm to serialize/deserialize the given binary tree. You only have to ensure that the serialized string can be deserialized to the original binary tree.

Note :

Serialization is the process of translating a data structure or object state into a format that can be stored or transmitted (for example, across a computer network) and reconstructed later. The opposite operation, that is, extracting a data structure from stored information, is deserialization.

Input Format:

The only line of input contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes.

Input format explanation:

The level order input for the tree depicted in the below image will be:

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation:
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level, and so on.
The input ends when all nodes at the last level are null(-1).

The above sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format :

Print the level order traversal of the deserialized binary tree separated by single spaces. For NULL nodes, print -1.

Note :

You don’t need to print anything; It has already been taken care of.

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Approaches

01 Approach

The idea is to store the level order traversal of the given binary tree to serialize the tree. But since only storing the level order traversal cannot uniquely determine the structure of the binary tree, we will insert -1 in place of NULL nodes so that we can determine the structure of the binary tree uniquely when we deserialize the binary tree.

We will store the serialized binary tree in a string, let’s say “serialized”.Also, we will use ‘,’ as a separator to separate the value of different nodes in the string because then we can distinguish that nodes 12 and 1,2 are different.

The steps to serialize the binary tree are as follows:

Define a queue; let’s say “level”.
Push the root of the given binary tree into the queue (“level”).
Now we will perform the following operations until the queue (“level”) does not become empty:
- Pop the node from the front of the “level’, let’s say this node is “curr”.
- If “curr” is not null, then append the value of “curr” to “serialized”, also append ‘,’ after this so that we can distinguish between values of different nodes. Also, we will push the left and right children of “curr” into the “level”.
- Otherwise if “curr” is null, then append “-1,” to “serialized”.
Return the “serialized” string.

To deserialize the tree, we will read the values in the “serialized” string one by one and then create the binary tree in a level order manner.

The steps to deserialize the binary tree are as follows:

Define a queue; let’s say “level”.
Read the value of the first node from the “serialized”, if this value is equal to -1, return a null node. Otherwise, create a new TreeNode, let’s say “root” with this value, and push this node into the “level”.
Now we will perform the following operations until “level” does not become empty:
- Pop the node at the front of “level’, let’s say this node is “curr”.
- Read the next two values from the “serialized” string. These will be the values of the left and the right child nodes of “curr”. Let’s say the two values are “leftChild” and “rightChild” respectively.
  - If “leftChild” is not equal to -1, then create a new TreeNode with the value “leftChild” and make this new node the left child of “curr” and push this node into “level”.
  - If “rightChild” is not equal to -1, then create a new TreeNode with the value “rightChild” and make this new node the right child of “curr” and push this node into “level”.
Return the root node.

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