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).