Leetcode 199. Binary Tree Right Side View

This problem is asking us to imagine looking at a binary tree from the right side - what nodes would be visible? At each horizontal level of the tree, only the rightmost node would be visible from this perspective.

We need to identify these rightmost nodes level by level, starting from the root (level 0) and going down to the deepest level. The challenge is efficiently determining which node is the rightmost at each level without having to examine every possible node.

The key insight is that 'rightmost' means the last node we encounter when traversing each level from left to right. This naturally suggests a level-by-level traversal approach.

The most straightforward approach would be to perform a complete level-order traversal of the entire tree, storing all nodes at each level in separate lists or arrays.

We would use BFS to traverse the tree level by level, collecting every single node at each depth. For each level, we'd maintain a list of all node values encountered during that level's traversal.

Once we've collected all nodes for a particular level, we'd simply take the last element from that level's list (since we traverse left to right, the last element is the rightmost). We'd repeat this process for every level until we've processed the entire tree.

This approach requires storing all nodes at each level temporarily, even though we only need the rightmost one, making it inefficient in terms of space usage.

from collections import deque
def rightSideView(root):
    if not root:
        return []
    
    result = []
    queue = deque([root])
    
    while queue:
        level_size = len(queue)
        level_nodes = []  # Store ALL nodes at this level
        
        # Process all nodes at current level
        for i in range(level_size):
            node = queue.popleft()
            level_nodes.append(node.val)  # Store every node value
            
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        
        # Take the rightmost (last) node from this level
        rightmost_value = level_nodes[-1]
        result.append(rightmost_value)
    
    return result

The breakthrough insight is that we don't need to store or examine all nodes at a level - we just need to process them in order and remember the last one we see at each level.

By tracking the queue size at the start of each level, we know exactly how many nodes belong to the current level, allowing us to process them completely before moving to the next level.

After seeing the brute force limitations of storing entire levels, level-order traversal with BFS becomes the perfect approach because it processes nodes level by level in the exact order we need.

BFS using a queue naturally groups nodes by their depth level. When we process all nodes at level N, we automatically enqueue all nodes for level N+1. This level-by-level processing is exactly what we need to identify the rightmost node at each level.

The queue's FIFO property ensures we process nodes from left to right within each level, making the last node we process at each level the rightmost node by definition.

Now that we understand why level-order traversal is perfect, here's how we implement it efficiently using BFS with a queue.

We'll use a queue to process nodes level by level, but instead of storing all nodes at each level, we'll track only the rightmost node as we traverse each level.

The key insight is to process each level completely before moving to the next. We determine the number of nodes at the current level (queue size), then process exactly that many nodes. As we process each node in the current level, we keep updating our 'rightmost' variable - the last node we process in each level will naturally be the rightmost.

For each level: dequeue all nodes at that level, add their children to the queue for the next level, and capture the value of the last node processed. This gives us the rightmost node at each level without storing unnecessary intermediate results.

from collections import deque
def rightSideView(root):
    if not root:
        return []
    
    result = []
    queue = deque([root])
    
    while queue:
        level_size = len(queue)
        rightmost = None
        
        # Process all nodes at current level
        for i in range(level_size):
            node = queue.popleft()
            rightmost = node.val
            
            # Add children for next level
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        
        # Add rightmost node value of current level
        result.append(rightmost)
    
    return result

• Level-order traversal (BFS) pattern: When you need to process a binary tree level by level, use a queue-based BFS approach - it naturally groups nodes by depth and processes them left-to-right, making it perfect for level-dependent problems like finding rightmost nodes.
• Last element extraction technique: Instead of tracking rightmost nodes during traversal, process each level completely and simply take the last element from each level - this eliminates the need for complex right-first traversal logic and works regardless of tree structure.
• Queue size snapshot optimization: Capture `queue.size()` at the start of each level and use it as your loop boundary - this ensures you process exactly one level at a time even as you add the next level's nodes to the queue during iteration.
• Empty tree edge case: Always check for `root == null` before starting BFS traversal - attempting to add null to a queue or access properties of null nodes will cause runtime errors, and empty trees should return empty lists.
• Level-based tree processing pattern: This BFS approach extends to any problem requiring level-wise analysis - finding level averages, detecting level-order patterns, or identifying nodes at specific depths all use the same queue-based level processing framework.