Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Move from brute-force thinking to an efficient approach using tree strategy.
Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Example 1:
Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]
Explanation:
Example 2:
Input: root = [1,2,3,4,null,null,null,5]
Output: [1,3,4,5]
Explanation:
Example 3:
Input: root = [1,null,3]
Output: [1,3]
Example 4:
Input: root = []
Output: []
Constraints:
[0, 100].-100 <= Node.val <= 100Problem summary: Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[1,2,3,null,5,null,4]
[1,2,3,4,null,null,null,5]
[1,null,3]
populating-next-right-pointers-in-each-node)boundary-of-binary-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #199: Binary Tree Right Side View
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
ans.add(q.peekFirst().val);
for (int k = q.size(); k > 0; --k) {
TreeNode node = q.poll();
if (node.right != null) {
q.offer(node.right);
}
if (node.left != null) {
q.offer(node.left);
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #199: Binary Tree Right Side View
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func rightSideView(root *TreeNode) (ans []int) {
if root == nil {
return
}
q := []*TreeNode{root}
for len(q) > 0 {
ans = append(ans, q[0].Val)
for k := len(q); k > 0; k-- {
node := q[0]
q = q[1:]
if node.Right != nil {
q = append(q, node.Right)
}
if node.Left != nil {
q = append(q, node.Left)
}
}
}
return
}
# Accepted solution for LeetCode #199: Binary Tree Right Side View
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
ans = []
if root is None:
return ans
q = deque([root])
while q:
ans.append(q[0].val)
for _ in range(len(q)):
node = q.popleft()
if node.right:
q.append(node.right)
if node.left:
q.append(node.left)
return ans
// Accepted solution for LeetCode #199: Binary Tree Right Side View
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut ans = vec![];
if root.is_none() {
return ans;
}
let mut q = VecDeque::new();
q.push_back(root);
while !q.is_empty() {
let k = q.len();
ans.push(q[0].as_ref().unwrap().borrow().val);
for _ in 0..k {
if let Some(node) = q.pop_front().unwrap() {
let mut node = node.borrow_mut();
if node.right.is_some() {
q.push_back(node.right.take());
}
if node.left.is_some() {
q.push_back(node.left.take());
}
}
}
}
ans
}
}
// Accepted solution for LeetCode #199: Binary Tree Right Side View
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function rightSideView(root: TreeNode | null): number[] {
const ans: number[] = [];
if (!root) {
return ans;
}
const q: TreeNode[] = [root];
while (q.length > 0) {
ans.push(q[0].val);
const nq: TreeNode[] = [];
for (const { left, right } of q) {
if (right) {
nq.push(right);
}
if (left) {
nq.push(left);
}
}
q.length = 0;
q.push(...nq);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.