Serialize and Deserialize Binary Tree in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Serialization with Preorder Traversal

What is the output of the serialization function when applied to this binary tree?

Tree structure:
1
/ \
2 3
/ \
4 5

The serialization uses preorder traversal with '#' for null nodes and commas as separators.

DSA Go

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func serialize(root *TreeNode) string {
    if root == nil {
        return "#"
    }
    leftSerialized := serialize(root.Left)
    rightSerialized := serialize(root.Right)
    return fmt.Sprintf("%d,%s,%s", root.Val, leftSerialized, rightSerialized)
}

// Tree construction
root := &TreeNode{1, &TreeNode{2, nil, nil}, &TreeNode{3, &TreeNode{4, nil, nil}, &TreeNode{5, nil, nil}}}

fmt.Println(serialize(root))

A 1,2,#,#,3,4,#,#,5,#,#

B 1,2,#,#,3,4,#,#,5,#

C 1,2,#,3,4,#,#,5,#,#

D 1,2,#,#,3,4,#,5,#,#,#

💡 Hint

Remember preorder traversal visits root, then left subtree, then right subtree. Null children are marked with '#'.

The serialization visits nodes in preorder: root, left, right. Null children are represented by '#'. So the output is "1,2,#,#,3,4,#,#,5,#,#".

❓ Predict Output

intermediate

Result of Deserializing a Serialized String

Given the serialized string "1,2,#,#,3,#,4,#,#", what is the value of the right child of the root after deserialization?

DSA Go

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func deserialize(data string) *TreeNode {
    vals := strings.Split(data, ",")
    var index int
    var build func() *TreeNode
    build = func() *TreeNode {
        if vals[index] == "#" {
            index++
            return nil
        }
        val, _ := strconv.Atoi(vals[index])
        index++
        node := &TreeNode{Val: val}
        node.Left = build()
        node.Right = build()
        return node
    }
    return build()
}

root := deserialize("1,2,#,#,3,#,4,#,#")
fmt.Println(root.Right.Val)

A 0

B 4

C 3

D nil (no right child)

💡 Hint

Trace the deserialization step by step, matching preorder nodes and nulls.

The root is 1, left child is 2 with no children, right child is 3 with no left child and right child 4. So root.Right.Val is 3.

🧠 Conceptual

advanced

Understanding Null Markers in Serialization

Why is it necessary to include a special marker (like '#') for null nodes when serializing a binary tree using preorder traversal?

A To indicate the end of the serialized string

B To distinguish between left and right children during deserialization

C To reduce the size of the serialized string

D To mark nodes that have only one child

💡 Hint

Think about how the deserialization function knows when to stop building a subtree.

Null markers tell the deserialization function where a child is missing, so it can correctly reconstruct the tree structure and distinguish left from right children.

❓ Predict Output

advanced

Output of Deserialization Followed by Serialization

What is the output of serializing the tree obtained by deserializing the string "1,#,2,#,3,#,#" using the given serialize and deserialize functions?

DSA Go

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func serialize(root *TreeNode) string {
    if root == nil {
        return "#"
    }
    leftSerialized := serialize(root.Left)
    rightSerialized := serialize(root.Right)
    return fmt.Sprintf("%d,%s,%s", root.Val, leftSerialized, rightSerialized)
}

func deserialize(data string) *TreeNode {
    vals := strings.Split(data, ",")
    var index int
    var build func() *TreeNode
    build = func() *TreeNode {
        if vals[index] == "#" {
            index++
            return nil
        }
        val, _ := strconv.Atoi(vals[index])
        index++
        node := &TreeNode{Val: val}
        node.Left = build()
        node.Right = build()
        return node
    }
    return build()
}

root := deserialize("1,#,2,#,3,#,#")
fmt.Println(serialize(root))

A 1,#,2,#,#

B 1,#,2,#,3,#

C 1,#,2,#,3,#,#,#

D 1,#,2,#,3,#,#

💡 Hint

The deserialized tree is skewed to the right. Serialization should reproduce the same string.

Deserializing then serializing returns the same preorder string with null markers, preserving the tree shape exactly.

🚀 Application

expert

Count Nodes from Serialized String Without Full Deserialization

Given a serialized binary tree string using preorder traversal with '#' for nulls, which approach correctly counts the total number of non-null nodes without building the tree?

A Use a stack to simulate preorder traversal, increment count for each number, and skip '#' markers

B Count all values that are not '#' by splitting the string by commas

C Count the number of commas and subtract the number of '#' markers

D Traverse the string and increment count only when a number is followed by two '#' markers

💡 Hint

Think about how preorder traversal works and how nulls affect node counting.

Using a stack to simulate traversal allows counting nodes correctly without building the tree. Simply counting non-'#' values may overcount if not careful.

Practice

1. What is the main purpose of serialize and deserialize functions for a binary tree?

easy

A. Convert a binary tree into a string and back to a tree

B. Sort the nodes of a binary tree in ascending order

C. Calculate the height of a binary tree

D. Delete all nodes in a binary tree

Solution

1. Step 1: Understand serialization

   Serialization means converting a binary tree into a string format so it can be stored or sent.

2. Step 2: Understand deserialization

   Deserialization means converting the string back into the original binary tree structure.

3. Final Answer:

   Convert a binary tree into a string and back to a tree -> Option A

4. Quick Check:

   Serialize/Deserialize = Convert tree to string and back [OK]

Hint: Serialization means tree to string, deserialization reverses it [OK]

Common Mistakes:

• Confusing serialization with sorting
• Thinking serialization deletes nodes
• Mixing serialization with tree height calculation

2. Which of the following Go function signatures correctly represents a deserialize function for a binary tree?

easy

A. func deserialize(root *TreeNode) string

B. func deserialize(data int) *TreeNode

C. func deserialize(data string) *TreeNode

D. func deserialize(data []int) int

Solution

1. Step 1: Identify input and output types

   Deserialize takes a string (serialized data) and returns a pointer to the root node of the tree.

2. Step 2: Match function signature

   func deserialize(data string) *TreeNode matches this: input is string, output is *TreeNode pointer.

3. Final Answer:

   func deserialize(data string) *TreeNode -> Option C

4. Quick Check:

   Deserialize input string, output tree node pointer [OK]

Hint: Deserialize input is string, output is tree root pointer [OK]

Common Mistakes:

• Using int instead of string as input
• Returning string instead of tree node
• Confusing serialize and deserialize signatures

3. Given the following serialized string of a binary tree using preorder with '#' as null marker: "1,2,#,#,3,4,#,#,5,#,#", what is the inorder traversal output of the deserialized tree?

medium

A. [1, 3, 2, 4, 5]

B. [1, 2, 3, 4, 5]

C. [2, 4, 1, 3, 5]

D. [2, 1, 4, 3, 5]

Solution

1. Step 1: Reconstruct tree from preorder string

   Preorder: 1 (root), 2 (left child), # (left null), # (right null), 3 (right child), 4 (left), #, #, 5 (right), #, #.

2. Step 2: Draw tree and find inorder traversal

   Tree structure:
   1
   / \
   2 3
   / \
   4 5
   Inorder (left, root, right): 2, 1, 4, 3, 5

3. Final Answer:

   [2, 1, 4, 3, 5] -> Option D

4. Quick Check:

   Inorder traversal = [2,1,4,3,5] [OK]

Hint: Draw tree from preorder, then do left-root-right for inorder [OK]

Common Mistakes:

• Mixing preorder and inorder traversal orders
• Ignoring null markers '#' in deserialization
• Incorrectly placing children nodes

4. Consider this Go code snippet for deserializing a binary tree. What is the bug?

func deserialize(data string) *TreeNode {
    vals := strings.Split(data, ",")
    var index int
    var build func() *TreeNode
    build = func() *TreeNode {
        if vals[index] == "#" {
            index++
            return nil
        }
        node := &TreeNode{Val: vals[index]}
        index++
        node.Left = build()
        node.Right = build()
        return node
    }
    return build()
}

medium

A. TreeNode.Val should be converted from string to int

B. Index variable should be global, not local

C. Strings.Split should use space instead of comma

D. The build function should not be recursive

Solution

1. Step 1: Check TreeNode.Val assignment

   vals[index] is a string, but TreeNode.Val expects an int. Conversion is missing.

2. Step 2: Identify fix

   Use strconv.Atoi(vals[index]) to convert string to int before assigning.

3. Final Answer:

   TreeNode.Val should be converted from string to int -> Option A

4. Quick Check:

   String to int conversion missing [OK]

Hint: Convert string to int before assigning to node value [OK]

Common Mistakes:

• Forgetting to convert string to int
• Changing index scope unnecessarily
• Misusing Split delimiter

5. You want to serialize a binary tree so that it can be stored and later deserialized exactly as it was. Which approach best handles trees with duplicate values and null nodes?

hard

A. Use inorder traversal without null markers

B. Use preorder traversal with a special marker for null nodes

C. Use postorder traversal ignoring null nodes

D. Store only node values in a list without structure

Solution

1. Step 1: Understand serialization needs

   To reconstruct the exact tree, we must record node order and null positions.

2. Step 2: Evaluate options

   Preorder with null markers records structure and values including duplicates. Inorder or postorder without nulls loses structure. Storing only values loses tree shape.

3. Final Answer:

   Use preorder traversal with a special marker for null nodes -> Option B

4. Quick Check:

   Preorder + null markers preserves structure and duplicates [OK]

Hint: Preorder + null markers saves full tree structure [OK]

Common Mistakes:

• Ignoring null nodes in serialization
• Using inorder without nulls loses tree shape
• Assuming values alone define tree