0199. Binary Tree Right Side View.md

BFS

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        // Edge case: empty tree
        if (!root) return {};

        vector<int> res;
        queue<TreeNode*> q;

        // Start BFS from the root
        q.push(root);

        while (!q.empty()) {
            int size = q.size();  // Number of nodes in the current level

            for (int i = 0; i < size; i++) {
                TreeNode* node = q.front();
                q.pop();

                // Push children for the next level
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);

                // The last node in this level is the rightmost one
                if (i == size - 1) {
                    res.push_back(node->val);
                }
            }
        }
        return res;
    }
};

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Problem Summary

Given the root of a binary tree, imagine standing on the right side of the tree.
Return the values of the nodes you can see, ordered from top to bottom.

This is a level-based visibility problem.

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Core Idea (BFS – Level Order Traversal)

We traverse the tree level by level using Breadth-First Search (BFS).

Key observation:

From the right side, only the last node in each level is visible.

So for each level:

1. Process all nodes in that level.
2. Record the value of the last node popped from the queue.

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Why This Works

• BFS guarantees nodes are processed level by level
• At each level, nodes are visited from left to right
• The last node processed in a level is exactly what you would see from the right side

This avoids unnecessary position tracking (such as column indices).

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Time Complexity

• O(n)
  Each node is visited exactly once.

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Space Complexity

• O(w) where w is the maximum width of the tree
  (Queue stores at most one full level at a time)

Worst case (complete binary tree): O(n)

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Key Takeaway

• This problem is about levels, not coordinates
• BFS + “take the last node of each level” is:
  • Simple
  • Intuitive
  • Interview-safe
• A classic example of using level-order traversal to extract structured information from a tree

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