Right Side View of Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result slice for storing the right side view nodes.

DSA Go

result := [1]

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

Anil

Bmap[int]int{}

Cmake([]int, 0, 10)

D[]int{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using a map instead of a slice.

Initializing with nil instead of an empty slice literal.

We use an empty slice of integers to store the right side view nodes dynamically.

2fill in blank

medium

Complete the code to append the rightmost node's value at each level to the result slice.

DSA Go

result = append(result, [1].Val)

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Anode

Broot

Ccurrent

Dqueue[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' which is always the tree root.

Using 'queue[0]' which may not be the current node.

We append the value of the current node being processed at the rightmost position of the level.

3fill in blank

hard

Fix the error in the loop condition to process nodes level by level.

DSA Go

for len([1]) > 0 {

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Astack

Bqueue

Cnodes

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Checking length of result which is unrelated to loop control.

Using stack which is not used here.

The queue holds nodes to process; loop continues while queue is not empty.

4fill in blank

hard

Fill both blanks to correctly enqueue the left child of the current node.

DSA Go

if [1] != nil {
    queue = append(queue, [2])
}

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Acurrent.Left

Bcurrent.Right

Attempts:

3 left

💡 Hint

Common Mistakes

Appending right child when checking left child.

Appending nil nodes.

We check if the left child exists and then append it to the queue.

5fill in blank

hard

Fill all three blanks to enqueue the right child after the left child inside the loop.

DSA Go

if [1] != nil {
    queue = append(queue, [2])
}
if [3] != nil {
    queue = append(queue, current.Right)
}

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

Acurrent.Left

Ccurrent.Right

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing left and right child checks.

Appending nil children.

First check and append left child, then check and append right child to the queue.

Practice

1. What does the "Right Side View of Binary Tree" represent?

easy

A. The nodes visible when the tree is viewed from the right side

B. The nodes visible when the tree is viewed from the left side

C. The nodes at the bottom level of the tree

D. The nodes at the top level of the tree

Solution

1. Step 1: Understand the view perspective

   The right side view shows nodes visible when looking at the tree from the right side.

2. Step 2: Compare with other views

   Left side view shows nodes from left, bottom view shows lowest nodes, top view shows highest nodes.

3. Final Answer:

   The nodes visible when the tree is viewed from the right side -> Option A

4. Quick Check:

   Right side view = nodes visible from right [OK]

Hint: Right side view shows last node at each level [OK]

Common Mistakes:

• Confusing right side view with left side view
• Thinking it shows bottom or top nodes only
• Assuming it shows all nodes at each level

2. Which Go code snippet correctly declares a binary tree node struct for the right side view problem?

easy

A. type TreeNode struct { Val int; Left, Right int }

B. type TreeNode struct { Value int; Left, Right TreeNode }

C. type TreeNode struct { Val int; Left, Right *TreeNode }

D. type TreeNode struct { Val int; Left *int; Right *int }

Solution

1. Step 1: Check struct field types

   Tree nodes must have pointers to left and right child nodes, so *TreeNode is correct.

2. Step 2: Validate field names and types

   Val is int, Left and Right are pointers to TreeNode, matching type TreeNode struct { Val int; Left, Right *TreeNode }.

3. Final Answer:

   type TreeNode struct { Val int; Left, Right *TreeNode } -> Option C

4. Quick Check:

   Pointers to child nodes = *TreeNode [OK]

Hint: Use pointers for child nodes in struct [OK]

Common Mistakes:

• Using non-pointer types for children
• Incorrect field names or types
• Using int instead of *TreeNode for children

3. What is the output of the right side view function for this tree?

    1
   / \
  2   3
   \   \
    5   4

Assuming the function returns a slice of node values visible from the right side.

medium

A. [1, 2, 4]

B. [1, 2, 5]

C. [1, 3, 5]

D. [1, 3, 4]

Solution

1. Step 1: Identify nodes visible from right at each level

   Level 1: 1; Level 2: 3 (right child of 1); Level 3: 4 (right child of 3)

2. Step 2: Collect these nodes in order

   The right side view is [1, 3, 4].

3. Final Answer:

   [1, 3, 4] -> Option D

4. Quick Check:

   Rightmost nodes per level = [1, 3, 4] [OK]

Hint: Pick last node at each level from right [OK]

Common Mistakes:

• Choosing left child nodes instead of right
• Including nodes not visible from right
• Mixing node order in output

4. Given this Go function snippet for right side view, what is the bug?

func rightSideView(root *TreeNode) []int {
    var result []int
    var queue []*TreeNode
    queue = append(queue, root)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if i == size-1 {
                result = append(result, node.Val)
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    return result
}

medium

A. The queue is not initialized properly

B. The function does not check if root is nil before appending

C. The loop condition should be i <= size

D. The function appends left child before right child, so right side view is incorrect

Solution

1. Step 1: Check root nil condition

   The function appends root to queue without checking if root is nil, which can cause runtime error.

2. Step 2: Verify other parts

   Queue initialization and loop conditions are correct; appending left before right is valid for BFS.

3. Final Answer:

   The function does not check if root is nil before appending -> Option B

4. Quick Check:

   Check root nil before use [OK]

Hint: Always check root != nil before BFS queue append [OK]

Common Mistakes:

• Ignoring nil root check causing panic
• Misunderstanding BFS queue order
• Incorrect loop boundary conditions

5. How would you modify the right side view algorithm to return the left side view instead?

hard

A. Change the condition to append node.Val when i == 0 instead of i == size-1

B. Change the order of child node appending to queue: append left child before right child

C. Change the condition to append node.Val when i == size-1 and reverse the queue

D. Change the order of child node appending to queue: append right child before left child

Solution

1. Step 1: Understand right side view logic

   Right side view appends node.Val when i == size-1 (last node at level).

2. Step 2: Modify for left side view

   Left side view requires first node at each level, so append when i == 0.

3. Final Answer:

   Change the condition to append node.Val when i == 0 instead of i == size-1 -> Option A

4. Quick Check:

   Left side view = first node per level [OK]

Hint: Append first node at each level for left side view [OK]

Common Mistakes:

• Only changing child append order without condition change
• Appending last node instead of first for left view
• Reversing queue unnecessarily