Interview Question for SDE-3s


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
1
of 1 vote

I am assuming the question was referring a BST. You can store the three basically in pre-order, in-order and post-order.
If you store the Tree in an in-order fashion you are basically storing the values sorted, in that case you can modify the file without having to rewrite the tree again.
Example:

BST:
           6
       3      7
    1    4      8

File:
1 3 4 6 7 8

Let's say you add 2 on the tree in order to update the file you only have to iterate until the number three and add it before.

- Fernando June 19, 2017 | Flag Reply
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0
of 0 vote

Serialization is to store tree in a file so that it can be later restored. The structure of tree must be maintained. Deserialization is reading tree back from file.

for more see Serialize and Deserialize a Binary Tree in geeksforgeeks

- kuppu612 June 18, 2017 | Flag Reply
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0
of 0 vote

I gave the same answer, But the ineterviewer was specific about not wirting the whole serialized file but writing the exactly changed node.

- Ghosh June 18, 2017 | Flag Reply
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0
of 0 vote

For the second part, if we first update and find the solution (in main memory) and then write only the parts that have been modified by traversing through the entire file. Would that be a possible solution ?

- kr0821 June 18, 2017 | Flag Reply
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0
of 0 vote

Well the question is what is changed, if there is a key / value pair, just locate the node and overwrite the value. If there's only one value that defines the trees structure, assuming it's a BST, this will always affect a large part of the tree. B-tree might be a solution then.

- ChrisK June 19, 2017 | Flag Reply
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0
of 0 vote

import java.io.*;

public class SerializeAndDeserialize {

    public static void serialize(BufferedWriter writer, Node root) throws IOException {
        if (root == null) return;
        writer.write(root.value);
        if (root.left == null && root.right == null){writer.write('*');writer.write('*'); return;}
        if (root.left == null) writer.write('*'); else serialize(writer, root.left);
        if (root.right == null) writer.write('*'); else serialize(writer, root.right);
    }

    public static Node deserialize(BufferedReader reader) throws IOException {
        int value = reader.read();
        if (value < 0 || value == '*') return null;
        Node temp = new Node (value);
        temp.left = deserialize(reader);
        temp.right = deserialize(reader);
        return temp;
    }

- noname June 21, 2017 | Flag Reply


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