What is the brute force approach and complexity for Binary Tree Right Side View?

Store each full level in a list, then take the back of each list with time complexity O(N) and space complexity O(N).

How do you identify the pattern for Binary Tree Right Side View?

Look for key signals in the prompt: "look at the tree from the right side" indicates Only one node per level is visible, so the answer has one value per depth.; "values from top to bottom" indicates The answer is in level order, so a level-by-level walk produces it directly.; "nodes you can see" indicates A node hidden by a node further right on its level is not counted. Only the rightmost node of each level matters..

Binary Tree Right Side View

Medium·LeetCode #199 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given the root of a binary tree.

Imagine you stand on the right side of the tree and look at it. Return the values of the nodes you can see, ordered from the top level down to the bottom level.

A node is visible when nothing on its level sits to its right. A node can be visible even if it is a left child.

Input

The root of a binary tree. The tree may be empty.

Output

An array of the visible node values, from top to bottom.

Example 1

Input: root = [1, 2, 3, null, 5, null, 4]

Output: [1, 3, 4]

Level 0 shows 1. Level 1 shows 3, its right child. Level 2 has 5 (under the left child 2) and 4 (under the right child 3). 4 is further right, so it hides 5.

Example 2

Input: root = [1, 2, 3, 4, null, null, null, 5]

Output: [1, 3, 4, 5]

Nodes 4 and 5 are both in the left subtree. Nothing is to their right on their levels, so they are visible.

Example 3

Input: root = []

Output: []

An empty tree has nothing to see.

Constraints

• The number of nodes is between 0 and 100.
• −100 ≤ Node.val ≤ 100

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“look at the tree from the right side”

→ Only one node per level is visible, so the answer has one value per depth.

“values from top to bottom”

→ The answer is in level order, so a level-by-level walk produces it directly.

“nodes you can see”

→ A node hidden by a node further right on its level is not counted. Only the rightmost node of each level matters.

Approach Complexity Ladder

Brute Force — keep every level, then read the last oneO(N)O(N)

Store each full level in a list, then take the back of each list

Better — BFS, keep the last node of each roundO(N)O(W)

Level-order traversal; the last node popped in each round goes into the answer

Optimal — DFS, right child firstO(N)O(H)

Visit the right subtree before the left; the first node reached at each depth is the answer for that depth

Brute Force Intuition

Run a normal level-order traversal, but save every level as a whole list. When the traversal ends, the answer is the last value of each list.

Recognition rule for this sub-pattern:

Compute something per level: last node (right view), max, average, width.

Remember this · Tree BFS

Trigger

Level order, per-level max/average, right side view, minimum depth, nodes at distance k.

Think

Does the question care about depth / rows rather than paths?

Key question

How many nodes are in the current level (queue size at the start of the level)?

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