Manufacturing Anxiety: Breaking the Illusion
This time in a Microsoft Coding Interview, the candidate encountered a classic binary tree problem.
Many people see "Binary Tree Right Side View" and think it's simple: "Isn't it just BFS? Take the last node from each level!"
Not that simple.
The focus of this problem is whether your understanding of level-order traversal is solid, and whether you can clearly explain the "level" boundaries during implementation. Microsoft isn't checking "do you know BFS" — they're checking whether you can deliver a complete, stable, unambiguous solution logic.
Problem Breakdown: Demonstrating Expertise
🔍 Original Problem (Interviewer's Exact Words)
The interviewer's description was concise:
You are given the root of a binary tree.
Imagine yourself standing on the right side of the tree.
Return the values of the nodes you can see ordered from top to bottom.
If the tree is empty, return an empty array.
After giving the problem, the interviewer didn't add any hints and directly asked the candidate to explain their understanding.
🎯 Example Analysis
Example 1:
Input:
1
/ \
2 3
\ \
5 4
Output: [1, 3, 4]
Explanation:
- Level 1: Only 1, right view sees 1
- Level 2: 2 and 3, right view sees 3 (rightmost)
- Level 3: 5 and 4, right view sees 4 (rightmost)
Example 2:
Input:
1
/
2
Output: [1, 2]
Explanation:
- Level 1: Only 1
- Level 2: Only 2 (though on left, it's the only node on this level, so visible)
Example 3:
Input: null
Output: []
Clarification Phase
The candidate didn't rush to explain the algorithm. Instead, they first made two confirmations — this step is crucial:
- "So each level only needs to output one value, right? The rightmost node of that level."
- "If the input root is null, just return an empty array, correct?"
The interviewer confirmed both understandings were correct.
This step is key because it directly defines the problem as: process by level + take only one node per level. All subsequent implementation revolves around this definition.
Deep Dive: Building Trust
🧠 Problem Essence Analysis
Core abilities tested:
| Test Point | Specific Content |
|---|---|
| BFS Level-Order Traversal | Use queue to process nodes level by level |
| Level Boundary Detection | Record current level's node count, finish before next level |
| Right Side View Definition | Last node processed in each level |
| Edge Case Handling | Empty tree, single node, left-skewed, right-skewed |
📊 Key Insights
Why BFS instead of DFS?
- BFS naturally processes by "level", clearly knowing which is the last node when each level ends
- DFS requires extra depth tracking and must traverse "right before left" to ensure correct right view
Level Boundary is Key
- In BFS, when entering the while loop, the queue contains exactly all nodes of the current level
- Record
levelSize = queue.size(), then for-loop processes exactly that many nodes - When
i == levelSize - 1, that's the last node of this level (right view node)
Solution: Core Algorithm
Solution: BFS Level-Order Traversal
from collections import deque
from typing import List, Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def rightSideView(root: Optional[TreeNode]) -> List[int]:
"""
Microsoft VO: Binary Tree Right Side View
Args:
root: Binary tree root node
Returns:
List[int]: List of node values visible from right side
"""
# Step 1: Edge case - empty tree returns empty array
if not root:
return []
result = []
queue = deque([root])
# Step 2: BFS level-order traversal
while queue:
# Record current level's node count (KEY!)
level_size = len(queue)
# Step 3: Process all nodes of current level
for i in range(level_size):
node = queue.popleft()
# Step 4: When processing last node of this level, record its value
if i == level_size - 1:
result.append(node.val)
# Step 5: Add left and right children to queue (left before right)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
Java Implementation (Common in Interviews)
import java.util.*;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int val) {
this.val = val;
}
}
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> result = new ArrayList<>();
// Step 1: Edge case
if (root == null) {
return result;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
// Step 2: BFS level-order traversal
while (!queue.isEmpty()) {
// Record current level's node count
int levelSize = queue.size();
// Step 3: Process all nodes of current level
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
// Step 4: Last node is the right view node
if (i == levelSize - 1) {
result.add(node.val);
}
// Step 5: Add children to queue
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
}
return result;
}
}
🧪 Test Cases
# Helper function to build tree from level-order array
def build_tree(values):
"""Build binary tree from level-order array"""
if not values or values[0] is None:
return None
root = TreeNode(values[0])
queue = deque([root])
i = 1
while queue and i < len(values):
node = queue.popleft()
if i < len(values) and values[i] is not None:
node.left = TreeNode(values[i])
queue.append(node.left)
i += 1
if i < len(values) and values[i] is not None:
node.right = TreeNode(values[i])
queue.append(node.right)
i += 1
return root
# Test 1: Standard example
root1 = build_tree([1, 2, 3, None, 5, None, 4])
print(rightSideView(root1)) # Expected: [1, 3, 4]
# Test 2: Left-skewed tree
root2 = build_tree([1, 2, None])
print(rightSideView(root2)) # Expected: [1, 2]
# Test 3: Empty tree
root3 = None
print(rightSideView(root3)) # Expected: []
# Test 4: Single node
root4 = TreeNode(1)
print(rightSideView(root4)) # Expected: [1]
# Test 5: Complete binary tree
root5 = build_tree([1, 2, 3, 4, 5, 6, 7])
print(rightSideView(root5)) # Expected: [1, 3, 7]
# Test 6: Right-skewed tree
root6 = build_tree([1, None, 2, None, 3])
print(rightSideView(root6)) # Expected: [1, 2, 3]
# Test 7: Uneven depth
root7 = build_tree([1, 2, 3, 4])
print(rightSideView(root7)) # Expected: [1, 3, 4]
Solution 2: DFS (Right Before Left)
def rightSideView_dfs(root: Optional[TreeNode]) -> List[int]:
"""
DFS solution: traverse right before left, first node at each depth is right view
"""
result = []
def dfs(node: TreeNode, depth: int):
if not node:
return
# If this depth is visited for first time, record this node
if depth == len(result):
result.append(node.val)
# Right before left traversal (ensures rightmost node visited first per level)
dfs(node.right, depth + 1)
dfs(node.left, depth + 1)
dfs(root, 0)
return result
Complexity Analysis
| Method | Time Complexity | Space Complexity |
|---|---|---|
| BFS Level-Order | O(n) | O(n) worst case queue stores one level |
| DFS Right-First | O(n) | O(h) recursion stack depth, h = tree height |
Where n = total nodes, h = tree height
🤯 Interviewer's Followup Traps
Followup 1: What about left side view?
Answer: Take the first node of each level instead of the last.
def leftSideView(root):
if not root:
return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
for i in range(level_size):
node = queue.popleft()
if i == 0: # First node is left view node
result.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
Followup 2: How to return all node values per level?
Answer: Classic "Level Order Traversal" (LeetCode 102).
def levelOrder(root):
if not root:
return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
level = []
for _ in range(level_size):
node = queue.popleft()
level.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(level)
return result
Followup 3: How about bottom view?
Answer: Need to track horizontal distance for each node, take the bottommost node at each horizontal distance.
Followup 4: What about vertical order traversal?
Answer: Group by horizontal distance, sort by depth within same horizontal distance.
🔥 Why Most People Fail
| Common Mistake | Correct Approach |
|---|---|
Not recording levelSize |
Record current level's node count when entering while loop |
Using queue.size() in for-loop condition |
Queue size changes, must fix it beforehand |
| Wrong order of adding children | Left before right (BFS) or right before left (DFS) |
| Not handling empty tree | First line should check null and return [] |
| Skipping clarification | Confirm "one value per level" and "null returns empty array" |
What the Interviewer Really Looks For
This problem isn't hard. What Microsoft checks in this round isn't "do you know BFS", but:
- Clarification Ability: Can you proactively clarify requirements and turn vague needs into clear definitions
- Expression Completeness: Can you deliver a complete, stable, unambiguous solution logic
- Implementation Stability: Can you write bug-free code under pressure
- Complexity Analysis: Can you accurately give time and space complexity
Code Template (Ready to Use)
from collections import deque
def rightSideView(root):
if not root:
return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
for i in range(level_size):
node = queue.popleft()
if i == level_size - 1:
result.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
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In this problem, oavoservice provided complete, ready-to-use solution text and implementation structure. The candidate just needed to deliver at the right pace to smoothly complete the entire problem.
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