Right Side View of Binary Tree in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Right Side View of a binary tree?

It is the list of nodes visible when looking at the tree from the right side. For each level, only the rightmost node is seen.

intermediate

How can we find the right side view of a binary tree using level order traversal?

Traverse the tree level by level (breadth-first). For each level, record the last node visited. These nodes form the right side view.

beginner

In Go, which data structure is commonly used to implement level order traversal for the right side view?

A queue is used to hold nodes of each level. Nodes are dequeued and their children enqueued to process level by level.

beginner

Why do we only keep the last node at each level for the right side view?

Because the last node at each level is the rightmost node visible from the right side. Nodes before it are hidden behind.

intermediate

What is the time complexity of finding the right side view of a binary tree using BFS?

The time complexity is O(n), where n is the number of nodes, because each node is visited once.

What does the right side view of a binary tree represent?

A All nodes at the deepest level

B Nodes visible from the left side, one per level

C All leaf nodes

D Nodes visible from the right side, one per level

The right side view shows the rightmost node at each level, visible when looking from the right.

Which traversal method is best suited to find the right side view of a binary tree?

A Preorder traversal

B Level order traversal (BFS)

C Inorder traversal

D Postorder traversal

Level order traversal processes nodes level by level, making it easy to pick the rightmost node at each level.

In Go, which data structure helps to perform level order traversal?

A Queue

B Map

C Stack

D Slice without order

A queue allows FIFO processing needed for level order traversal.

If a binary tree has 10 nodes, what is the time complexity to find its right side view using BFS?

A O(n)

B O(log n)

C O(1)

D O(n^2)

Each node is visited once, so time complexity is linear, O(n).

What node do we record at each level to form the right side view?

A First node at the level

B Middle node at the level

C Last node at the level

D All nodes at the level

The last node at each level is the rightmost and visible from the right side.

Explain how to find the right side view of a binary tree using a queue in Go.

Think about visiting nodes one level at a time and picking the rightmost node.

Describe why the right side view only includes one node per level and which node that is.

Imagine standing on the right side and looking at the tree.

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