Binary Tree Node Structure in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a binary tree node with an integer value.

DSA Go

type Node struct {
    Value int
    Left  *Node
    Right *[1]
}

Drag options to blanks, or click blank then click option'

ANodePtr

Bint

CTree

DNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a non-pointer type for child nodes causes incorrect tree structure.

Using unrelated types like int or Tree instead of Node.

The Left and Right fields should be pointers to Node to represent child nodes in the binary tree.

2fill in blank

medium

Complete the code to create a new node with a given integer value.

DSA Go

func NewNode(val int) *Node {
    return &Node{Value: [1]
}

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ANode

Bval

Cval.Value

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable like value or Node instead of val.

Trying to access val.Value which does not exist.

The function parameter val holds the integer value to assign to the new node's Value field.

3fill in blank

hard

Fix the error in the code to assign the left child node correctly.

DSA Go

func (n *Node) SetLeft(child *Node) {
    n.Left = [1]
}

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A&child

B*child

Cchild

DNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using &child causes type mismatch (pointer to pointer).

Using *child tries to assign a value instead of a pointer.

The Left field expects a pointer to Node, so assign it directly to child which is already a pointer.

4fill in blank

hard

Fill both blanks to check if a node is a leaf (no children).

DSA Go

func (n *Node) IsLeaf() bool {
    return n.Left == [1] && n.Right == [2]
}

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Anil

Btrue

Cfalse

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using true or false instead of nil for pointer checks.

Using 0 which is not valid for pointers.

A leaf node has no children, so both Left and Right pointers should be nil.

5fill in blank

hard

Fill all three blanks to create a map of node values to their left child values if present.

DSA Go

func LeftChildMap(root *Node) map[int]int {
    result := make(map[int]int)
    if root.Left != [1] {
        result[[2]] = root.Left.[3]
    }
    return result
}

Drag options to blanks, or click blank then click option'

Anil

Broot.Value

CValue

Droot.Left

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.Left instead of root.Value as map key.

Using root.Left instead of Value to get child's value.

Checking against true or false instead of nil.

Check if Left child is not nil, then map root's Value to Left child's Value field.