Tree Traversal Inorder Left Root Right in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the left subtree first in inorder traversal.

DSA Go

func inorder(node *Node) {
    if node == nil {
        return
    }
    inorder(node.[1])
    fmt.Print(node.value, " ")
    inorder(node.right)
}

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

Aroot

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Calling inorder on node.right first.

Calling inorder on node.parent which does not exist.

Skipping the left subtree.

In inorder traversal, we first visit the left child, then the root, then the right child. So we call inorder on node.left first.

2fill in blank

medium

Complete the code to print the root value in inorder traversal.

DSA Go

func inorder(node *Node) {
    if node == nil {
        return
    }
    inorder(node.left)
    fmt.Print(node.[1], " ")
    inorder(node.right)
}

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

Aleft

Bvalue

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing node.left or node.right instead of node.value.

Printing node.parent which is not part of the node struct.

In inorder traversal, after visiting the left subtree, we print the root node's value.

3fill in blank

hard

Fix the error in the inorder traversal code to correctly traverse the right subtree.

DSA Go

func inorder(node *Node) {
    if node == nil {
        return
    }
    inorder(node.left)
    fmt.Print(node.value, " ")
    inorder(node.[1])
}

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

Aright

Bleft

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Calling inorder on node.left again instead of node.right.

Calling inorder on node.parent or node.root which do not exist.

After printing the root, inorder traversal visits the right subtree, so we call inorder on node.right.

4fill in blank

hard

Fill both blanks to complete the inorder traversal function that prints nodes in left-root-right order.

DSA Go

func inorder(node *Node) {
    if node == nil {
        return
    }
    inorder(node.[1])
    fmt.Print(node.[2], " ")
    inorder(node.right)
}

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

Aleft

Bvalue

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children.

Printing node.left or node.right instead of node.value.

The first blank is 'left' to visit the left subtree first. The second blank is 'value' to print the node's data.

5fill in blank

hard

Fill all three blanks to complete the inorder traversal function that prints nodes in left-root-right order.

DSA Go

func inorder(node *Node) {
    if node == nil {
        return
    }
    inorder(node.[1])
    fmt.Print(node.[2], " ")
    inorder(node.[3])
}

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

Aleft

Bvalue

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.parent which does not exist.

Swapping left and right children in recursive calls.

The blanks are filled with 'left' to visit left subtree, 'value' to print node data, and 'right' to visit right subtree.

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Root node, Right subtree, Left subtree

B. Root node, Left subtree, Right subtree

C. Right subtree, Root node, Left subtree

D. Left subtree, Root node, Right subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal means visiting the left subtree first, then the root node, and finally the right subtree.

2. Step 2: Apply the traversal order

   Following this order ensures nodes are visited in ascending order for binary search trees.

3. Final Answer:

   Left subtree, Root node, Right subtree -> Option D

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder = Left, Root, Right [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Visiting root before left subtree
• Visiting right subtree before root

2. Which of the following Go function signatures correctly represents an inorder traversal function for a binary tree node defined as type Node struct { Val int; Left, Right *Node }?

easy

A. func inorder(root *Node) []int

B. func inorder(root Node) []int

C. func inorder(root *Node) int

D. func inorder(root Node) int

Solution

1. Step 1: Understand pointer usage in Go for trees

   Binary trees use pointers to nodes to allow recursive traversal and modification.

2. Step 2: Choose function signature returning slice of ints

   Inorder traversal returns a list of node values, so return type should be []int, and input should be pointer to Node.

3. Final Answer:

   func inorder(root *Node) []int -> Option A

4. Quick Check:

   Pointer receiver and slice return type [OK]

Hint: Use pointer receiver and return slice for traversal [OK]

Common Mistakes:

• Passing Node by value instead of pointer
• Returning single int instead of slice
• Using incorrect return type

3. What is the output of the inorder traversal for the following binary tree?

    2
   / \
  1   3

Given the Go code:

type Node struct {
    Val int
    Left, Right *Node
}

func inorder(root *Node) []int {
    if root == nil {
        return []int{}
    }
    left := inorder(root.Left)
    right := inorder(root.Right)
    return append(append(left, root.Val), right...)
}

func main() {
    root := &Node{2, &Node{1, nil, nil}, &Node{3, nil, nil}}
    result := inorder(root)
    fmt.Println(result)
}

medium

A. [3, 2, 1]

B. [2, 1, 3]

C. [1, 2, 3]

D. [1, 3, 2]

Solution

1. Step 1: Trace inorder traversal on given tree

   Visit left subtree (1), then root (2), then right subtree (3).

2. Step 2: Confirm output list order

   The traversal returns [1, 2, 3] as expected for inorder.

3. Final Answer:

   [1, 2, 3] -> Option C

4. Quick Check:

   Inorder traversal output = [1, 2, 3] [OK]

Hint: Inorder visits left, root, then right [OK]

Common Mistakes:

• Appending root before left subtree
• Appending right subtree before root
• Confusing preorder or postorder output

4. The following Go code attempts inorder traversal but produces incorrect output. What is the bug?

func inorder(root *Node) []int {
    if root == nil {
        return []int{}
    }
    left := inorder(root.Left)
    right := inorder(root.Right)
    return append(left, right...)
    return append(result, root.Val)
}

medium

A. The root value is appended after the return statement, so never executed

B. The function should return int, not []int

C. The base case should return nil, not empty slice

D. The left subtree is not traversed

Solution

1. Step 1: Analyze return statements in function

   The first return statement returns append(left, right...), so the second return is unreachable.

2. Step 2: Identify missing root value append

   Root value is never appended because code after first return is ignored.

3. Final Answer:

   The root value is appended after the return statement, so never executed -> Option A

4. Quick Check:

   Code after return is unreachable [OK]

Hint: Check for unreachable code after return [OK]

Common Mistakes:

• Ignoring unreachable code after return
• Not appending root value in correct order
• Confusing base case return value

5. Given a binary tree where some nodes have only one child, how does inorder traversal help in printing the nodes in sorted order if the tree is a binary search tree? Consider the tree:

      5
     / \
    3   8
     \   \
      4   10

What is the inorder traversal output?

hard

A. [5, 3, 4, 8, 10]

B. [3, 4, 5, 8, 10]

C. [4, 3, 5, 10, 8]

D. [3, 5, 4, 8, 10]

Solution

1. Step 1: Apply inorder traversal on given tree

   Visit left subtree: node 3, then its right child 4, then root 5, then right subtree nodes 8 and 10.

2. Step 2: List nodes in inorder sequence

   Sequence is 3, 4, 5, 8, 10 which is sorted ascending order.

3. Final Answer:

   [3, 4, 5, 8, 10] -> Option B

4. Quick Check:

   Inorder traversal of BST gives sorted nodes [OK]

Hint: Inorder traversal of BST prints nodes sorted [OK]

Common Mistakes:

• Ignoring nodes with single child
• Visiting root before left subtree
• Mixing preorder or postorder with inorder