It is the list of nodes visible when the tree is seen from the right side. Only the rightmost node at each level is included.
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beginner
Which traversal method is commonly used to find the right side view of a binary tree?
Level-order traversal (Breadth-First Search) is commonly used, visiting nodes level by level and recording the last node at each level.
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beginner
In the right side view algorithm, why do we record the last node at each level?
Because the last node visited at each level is the rightmost node visible from the right side.
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beginner
What data structure helps to implement the level-order traversal for right side view?
A queue is used to keep track of nodes at each level during the traversal.
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beginner
How does the right side view differ from the left side view of a binary tree?
Right side view shows the rightmost nodes at each level, while left side view shows the leftmost nodes at each level.
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What node do you record at each level to get the right side view?
AThe first node at each level
BAll nodes at each level
CThe root node only
DThe last node at each level
✗ Incorrect
The last node at each level is the rightmost node visible from the right side.
Which data structure is best for level-order traversal to find the right side view?
AQueue
BSet
CStack
DMap
✗ Incorrect
A queue helps process nodes level by level in breadth-first order.
If a binary tree has only left children, what will the right side view show?
AAll nodes in a straight line
BOnly the root node
CNo nodes
DOnly leaf nodes
✗ Incorrect
Since all nodes are on the left, the right side view will show all nodes as they appear from the right.
What is the time complexity of finding the right side view using level-order traversal?
AO(1)
BO(log n)
CO(n)
DO(n^2)
✗ Incorrect
Each node is visited once, so the time complexity is linear in the number of nodes.
Which of these is NOT true about the right side view of a binary tree?
AIt can be found using depth-first search
BIt includes all nodes at each level
CIt includes the rightmost node at each level
DIt shows what is visible from the right side
✗ Incorrect
The right side view includes only one node per level, not all nodes.
Explain how to find the right side view of a binary tree using level-order traversal.
Think about visiting nodes level by level and picking the rightmost one.
You got /4 concepts.
Describe the difference between right side view and left side view of a binary tree.
Consider which nodes are visible from each side.
You got /4 concepts.
Practice
(1/5)
1. What does the "Right Side View of Binary Tree" represent?
easy
A. Nodes visible when the tree is viewed from the right side
B. Nodes visible when the tree is viewed from the left side
C. All leaf nodes of the binary tree
D. All nodes at the deepest level of the tree
Solution
Step 1: Understand the view perspective
The right side view shows nodes visible when looking at the tree from the right side.
Step 2: Differentiate from other views
Unlike left side or leaf views, this specifically shows the rightmost nodes at each level.
Final Answer:
Nodes visible when the tree is viewed from the right side -> Option A
Quick Check:
Right side view = rightmost nodes [OK]
Hint: Right side view shows rightmost nodes at each level [OK]
Common Mistakes:
Confusing right side view with left side view
Thinking it shows all leaf nodes
Assuming it shows deepest level nodes only
2. Which of the following Python code snippets correctly initializes an empty list to store the right side view nodes?
easy
A. right_view = set()
B. right_view = []
C. right_view = ()
D. right_view = {}
Solution
Step 1: Identify the correct data structure for ordered collection
We need a list to store nodes in order, so an empty list is initialized with [].
Step 2: Check other options
{} is a dictionary, () is a tuple, set() is unordered; none fit the requirement.
Final Answer:
right_view = [] -> Option B
Quick Check:
Empty list uses [] [OK]
Hint: Use [] for empty list to store nodes [OK]
Common Mistakes:
Using {} which creates a dictionary
Using () which creates an empty tuple
Using set() which is unordered
3. What is the output of the following code that prints the right side view of the binary tree?
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def rightSideView(root):
if not root:
return []
from collections import deque
queue = deque([root])
right_view = []
while queue:
level_length = len(queue)
for i in range(level_length):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if i == level_length - 1:
right_view.append(node.val)
return right_view
# Tree:
# 1
# / \
# 2 3
# \ \
# 5 4
root = TreeNode(1, TreeNode(2, None, TreeNode(5)), TreeNode(3, None, TreeNode(4)))
print(rightSideView(root))
medium
A. [1, 3, 4]
B. [1, 2, 5]
C. [1, 3, 5]
D. [1, 2, 4]
Solution
Step 1: Traverse level by level and identify rightmost nodes
Level 1: Node 1 (only node), Level 2: Nodes 2 and 3, rightmost is 3, Level 3: Nodes 5 and 4, rightmost is 4.
Step 2: Collect rightmost nodes in order
Right side view list is [1, 3, 4].
Final Answer:
[1, 3, 4] -> Option A
Quick Check:
Rightmost nodes per level = [1, 3, 4] [OK]
Hint: Pick last node at each level during BFS [OK]
Common Mistakes:
Picking leftmost node instead of rightmost
Appending nodes before checking level end
Confusing node values at each level
4. Find the error in the following code snippet for right side view and select the fix:
def rightSideView(root):
if not root:
return []
queue = [root]
right_view = []
while queue:
level_length = len(queue)
for i in range(level_length):
node = queue.pop()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if i == level_length - 1:
right_view.append(node.val)
return right_view
medium
A. Remove the if condition checking i == level_length - 1
B. Replace queue.append(node.left) with queue.insert(0, node.left)
C. Initialize queue as a deque instead of list
D. Replace queue.pop() with queue.pop(0) to pop from front
Solution
Step 1: Identify queue operation error
Using pop() removes last element, but BFS requires removing from front (FIFO).
Step 2: Fix by popping from front
Use pop(0) to remove first element, maintaining correct BFS order.
Final Answer:
Replace queue.pop() with queue.pop(0) to pop from front -> Option D
Quick Check:
Queue pop front for BFS [OK]
Hint: Use pop(0) to dequeue from front in list-based BFS [OK]