Right Side View of Binary Tree in DSA Go - Execution Trace

Concept Flow - Right Side View of Binary Tree

Start at root node

↓

Use queue for level order traversal

↓

For each level:

↓

Traverse nodes left to right

↓

Record last node of level (rightmost)

↓

Add children to queue

↓

Repeat until queue empty

↓

Return list of recorded nodes

We visit the tree level by level from left to right, recording the last node at each level to get the right side view.

Execution Sample

DSA Go

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

This code collects the rightmost node at each level of the binary tree using a queue for level order traversal.

Execution Table

Step	Operation	Node Visited	Queue State	Rightmost Node This Level	Right Side View So Far
1	Start with root	1	[1]	None	[]
2	Process level size=1	1	[]	1	[1]
3	Add children of 1	1	[2,3]	1	[1]
4	Process level size=2	2	[3]	2	[1]
5	Add children of 2	2	[3,5]	2	[1]
6	Process level size=1	3	[5,4]	3	[1,3]
7	Add children of 3	3	[5,4]	3	[1,3]
8	Process level size=2	5	[4]	5	[1,3]
9	Add children of 5	5	[4]	5	[1,3]
10	Process level size=1	4	[]	4	[1,3,4]
11	Add children of 4	4	[]	4	[1,3,4]
12	Queue empty, end	None	[]	None	[1,3,4]

💡 Queue is empty, all levels processed, right side view collected.

Variable Tracker

Variable	Start	After Step 2	After Step 6	After Step 10	Final
queue	[1]	[2,3]	[5,4]	[]	[]
rightmostNode	None	1	3	4	None
rightSideView	[]	[1]	[1,3]	[1,3,4]	[1,3,4]

Key Moments - 3 Insights

Why do we record the last node visited at each level as the rightmost node?

Because we traverse nodes left to right at each level, the last node processed is the rightmost visible node from the right side. See execution_table rows 2, 6, and 10 where the last node is recorded.

Why do we use a queue for traversal instead of recursion?

A queue helps us process nodes level by level (breadth-first), which is necessary to identify the rightmost node at each level. This is shown in the queue state changes in the execution_table.

What happens if the tree is empty?

The function returns an empty list immediately (see execution_sample code line 2), because there are no nodes to view.

Visual Quiz - 3 Questions

Look at the execution_table at Step 6, what is the queue state?

A [2,3]

B [5,4]

C [3]

D []

💡 Hint

Check the 'Queue State' column at Step 6 in the execution_table.

At which step does the rightSideView first include the node '3'?

A Step 8

B Step 4

C Step 6

D Step 10

💡 Hint

Look at the 'Right Side View So Far' column and find when '3' appears.

If the tree had no right child for node 1, how would the rightSideView change?

A [1,2,5]

B [1,2]

C [1,2,4]

D [1,3,4]

💡 Hint

Without right child 3, the rightmost nodes come from left children only, check how queue and rightmost nodes change.

Concept Snapshot

Right Side View of Binary Tree:
- Use level order traversal (queue) to visit nodes level by level.
- At each level, record the last node visited (rightmost node).
- Add children left to right to queue for next level.
- Continue until queue is empty.
- Result is list of rightmost nodes per level.

Full Transcript

To find the right side view of a binary tree, we visit nodes level by level using a queue. At each level, we traverse nodes from left to right and record the last node visited as the rightmost visible node. We add children of each node to the queue to process the next level. This continues until all levels are processed and the queue is empty. The collected nodes form the right side view. If the tree is empty, we return an empty list immediately.

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