How do you solve Binary Tree Right Side View on LeetCode?

Binary Tree Right Side View (LeetCode #199) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using DFS allows us to capture the rightmost nodes efficiently.

What is the brute force solution for Binary Tree Right Side View?

The brute force approach for Binary Tree Right Side View is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing the entire tree and collecting the rightmost node at each level. This is straightforward but inefficient, as it may require multiple passes through the tree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Right Side View?

Binary Tree Right Side View uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Breadth-First Search.

What are the interview tips for Binary Tree Right Side View?

Always clarify the problem statement and constraints before diving into coding. Think about both iterative and recursive solutions; sometimes one is more intuitive than the other. Practice explaining your thought process as you code; it helps interviewers understand your approach.

What are common mistakes when solving Binary Tree Right Side View?

Not considering edge cases like empty trees. Confusing the order of traversal (left vs right) when trying to capture the rightmost nodes.

#199

Binary Tree Right Side View

Medium

TreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchBreadth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses depth-first search (DFS) to traverse the tree while keeping track of the current depth. This allows us to capture the first node we encounter at each depth, which will be the rightmost node.

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Algorithm

4 steps

1Step 1: Initialize a list to store the result and a helper function for DFS.
2Step 2: In the DFS function, check if the current depth is equal to the length of the result list.
3Step 3: If it is, append the current node's value to the result list.
4Step 4: Recursively call DFS for the right child first, then the left child.

solution.py11 lines

1def rightSideView(root):
2    result = []
3    def dfs(node, depth):
4        if not node:
5            return
6        if depth == len(result):
7            result.append(node.val)
8        dfs(node.right, depth + 1)
9        dfs(node.left, depth + 1)
10    dfs(root, 0)
11    return result

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the recursion stack in the worst case of a skewed tree.

1Using DFS allows us to capture the rightmost nodes efficiently.
2The depth of recursion helps us track which level we are on in the tree.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.