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DSA Goprogramming~10 mins

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

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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
StepOperationNode VisitedQueue StateRightmost Node This LevelRight Side View So Far
1Start with root1[1]None[]
2Process level size=11[]1[1]
3Add children of 11[2,3]1[1]
4Process level size=22[3]2[1]
5Add children of 22[3,5]2[1]
6Process level size=13[5,4]3[1,3]
7Add children of 33[5,4]3[1,3]
8Process level size=25[4]5[1,3]
9Add children of 55[4]5[1,3]
10Process level size=14[]4[1,3,4]
11Add children of 44[]4[1,3,4]
12Queue empty, endNone[]None[1,3,4]
💡 Queue is empty, all levels processed, right side view collected.
Variable Tracker
VariableStartAfter Step 2After Step 6After Step 10Final
queue[1][2,3][5,4][][]
rightmostNodeNone134None
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
Test your understanding
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'?
AStep 8
BStep 4
CStep 6
DStep 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.