Binary Tree Right Side View

Medium Trees & BST Views Original on LeetCode

Binary Tree Right Side View is a medium trees & bst problem solved with the views pattern. The best approach, dfs, right child first, runs in O(n) time and O(h) space. Below are 2 approaches in Java, from bfs, last node per level up.

Problem

Imagine standing on the right side of a binary tree. Return the values you can see, from top to bottom.

Examples

Example 1

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

Example 2

Input: root = [1, 2, 3, 4]
Output: [1, 3, 4]
Why: 4 is on the left, but nothing is to its right on that level.

Constraints

The tree has 0 to 100 nodes.

Animated walkthrough

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Code

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
BFS, last node per level	`O(n)`	`O(w)`
DFS, right child first	`O(n)`	`O(h)`

1BFS, last node per level

TimeO(n)

SpaceO(w)

Record the last node polled on each level.

For each level record the node at i == size - 1.

Java

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> out = new ArrayList<>();
        if (root == null) return out;
        Queue<TreeNode> q = new ArrayDeque<>();
        q.add(root);
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode n = q.poll();
                if (i == size - 1) out.add(n.val);
                if (n.left != null) q.add(n.left);
                if (n.right != null) q.add(n.right);
            }
        }
        return out;
    }
}

2DFS, right child first

TimeO(n)

SpaceO(h)

Preorder visiting right before left. The first node reached at each new depth is the rightmost on that level.

If depth == out.size(), add node.val.
Recurse right, then left.

Java

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> out = new ArrayList<>();
        dfs(root, 0, out);
        return out;
    }

    private void dfs(TreeNode n, int depth, List<Integer> out) {
        if (n == null) return;
        if (depth == out.size()) out.add(n.val);
        dfs(n.right, depth + 1, out);
        dfs(n.left, depth + 1, out);
    }
}

Edge cases to test

Empty tree
A deeper left subtree

Hints

Hint 1

The right view is the last node of each level, or the first node reached at each depth when DFS goes right first.

FAQ

What is the best time complexity for Binary Tree Right Side View?

DFS, right child first runs in O(n) time and O(h) extra space.

Which pattern does Binary Tree Right Side View use?

It is a trees & bst problem that uses the views pattern. Other problems with the same pattern: Top View of Binary Tree, Bottom View of Binary Tree, Left View of Binary Tree.

Is there a brute force solution for Binary Tree Right Side View?

Yes. BFS, last node per level takes O(n) time and O(w) space. Record the last node polled on each level.

Which edge cases should I test for Binary Tree Right Side View?

Empty tree; A deeper left subtree.