Right Side View of Binary Tree in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Right Side View for a Simple Binary Tree

What is the output of the right side view function for the given binary tree?

DSA Go

package main

import "fmt"

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

func rightSideView(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    var result []int
    queue := []*TreeNode{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
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Right = &TreeNode{Val: 4}
    fmt.Println(rightSideView(root))
}

A [1, 3, 4]

B [1, 2, 5]

C [1, 3, 5]

D [1, 2, 4]

💡 Hint

Think about which nodes are visible when looking from the right side level by level.

The right side view shows the last node at each level when traversed from left to right. At level 1: 1, level 2: 3, level 3: 4.

🧠 Conceptual

intermediate

Understanding the Right Side View Concept

Which statement best describes the right side view of a binary tree?

A It contains the leftmost node at each level of the binary tree.

B It contains all leaf nodes of the binary tree.

C It contains all nodes visible when the tree is viewed from the right side, showing the rightmost node at each level.

D It contains all nodes in the tree in preorder traversal order.

💡 Hint

Focus on what nodes are visible from the right side perspective.

The right side view includes only the nodes that are visible when looking at the tree from the right side, which means the rightmost node at each level.

🔧 Debug

advanced

Identify the Bug in Right Side View Implementation

What error will occur when running this code snippet for right side view of a binary tree?

DSA Go

func rightSideView(root *TreeNode) []int {
    if root == nil {
        return nil
    }
    var result []int
    queue := []*TreeNode{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
}

A runtime error: index out of range

B compiles and runs correctly

C returns empty slice

D syntax error due to missing colon

💡 Hint

Check the loop condition carefully for off-by-one errors.

The for loop uses i <= size which causes one extra iteration leading to accessing queue[0] when queue is empty, causing index out of range error.

🚀 Application

advanced

Right Side View for a Skewed Binary Tree

Given a binary tree where every node has only a left child, what will be the right side view output?

DSA Go

root := &TreeNode{Val: 1}
root.Left = &TreeNode{Val: 2}
root.Left.Left = &TreeNode{Val: 3}
root.Left.Left.Left = &TreeNode{Val: 4}
// Call rightSideView(root)

A [1]

B [1, 2, 3, 4]

C []

D [4]

💡 Hint

Think about which nodes are visible from the right side when the tree is skewed left.

Since there are no right children, the right side view will show all nodes as they appear from the right side, which is the same as the left skewed path.

❓ Predict Output

expert

Output of Right Side View for Complex Tree with Multiple Levels

What is the output of the rightSideView function for the following tree structure?

DSA Go

root := &TreeNode{Val: 10}
root.Left = &TreeNode{Val: 5}
root.Right = &TreeNode{Val: 15}
root.Left.Left = &TreeNode{Val: 3}
root.Left.Right = &TreeNode{Val: 7}
root.Right.Left = &TreeNode{Val: 12}
root.Right.Right = &TreeNode{Val: 18}
root.Left.Right.Right = &TreeNode{Val: 8}
root.Right.Right.Left = &TreeNode{Val: 17}
// Call rightSideView(root)

A [10, 15, 12, 17]

B [10, 15, 18, 8]

C [10, 5, 7, 8]

D [10, 15, 18, 17]

💡 Hint

Remember to pick the rightmost node at each level.

Level 1 rightmost: 10, level 2 rightmost: 15, level 3 rightmost: 18, level 4 rightmost: 17 (since 18 has left child 17).

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