Right Side View of Binary Tree in DSA Go - Time & Space Complexity

Time Complexity: Right Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the right side view of a binary tree changes as the tree grows.

Specifically, how does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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
}
    

This code uses a level-by-level traversal to collect the rightmost node's value at each level of the binary tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs while there are nodes to process, and the inner loop visits each node at the current level.
• How many times: Each node in the tree is visited exactly once during the traversal.

How Execution Grows With Input

As the number of nodes (n) in the tree increases, the code visits each node once, so the total steps grow roughly in direct proportion to n.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the right side view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The time complexity is O(log n) because we only look at one node per level."

[OK] Correct: Even though we record one node per level, we still visit every node to find that rightmost node, so the time depends on all nodes, not just the height.

Interview Connect

Understanding this traversal and its time complexity helps you explain how tree algorithms work efficiently, a common skill interviewers look for.

Self-Check

"What if we changed the traversal to a depth-first search instead of breadth-first? How would the time complexity change?"

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