Description
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:
- The number of nodes in the tree is in the range
[0, 100].
-100 <= Node.val <= 100
Solutions
Solution 1: BFS
We can use breadth-first search (BFS) and define a queue q to store the nodes. We start by putting the root node into the queue. Each time, we take out all the nodes of the current level from the queue. For the current node, we first check if the right subtree exists; if it does, we put the right subtree into the queue. Then, we check if the left subtree exists; if it does, we put the left subtree into the queue. This way, the first node taken out from the queue each time is the rightmost node of that level.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScriptRustJavaScript
# 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(code-box)
/**
* 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;
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> ans;
if (!root) {
return ans;
}
queue<TreeNode*> q{{root}};
while (q.size()) {
ans.push_back(q.front()->val);
for (int k = q.size(); k; --k) {
auto node = q.front();
q.pop();
if (node->right) {
q.push(node->right);
}
if (node->left) {
q.push(node->left);
}
}
}
return ans;
}
};(code-box)
/**
* 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
}(code-box)
/**
* 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;
}(code-box)
// 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
}
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var rightSideView = function (root) {
const ans = [];
if (!root) {
return ans;
}
const q = [root];
while (q.length > 0) {
ans.push(q[0].val);
const nq = [];
for (const { left, right } of q) {
if (right) {
nq.push(right);
}
if (left) {
nq.push(left);
}
}
q.length = 0;
q.push(...nq);
}
return ans;
};(code-box)
Solution 2: DFS
Use DFS (depth-first search) to traverse the binary tree. Each time, traverse the right subtree first, then the left subtree. This way, the first node visited at each level is the rightmost node of that level.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScriptJavaScriptRustJavaScript
# 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]:
def dfs(root: Optional[TreeNode], depth: int) -> None:
if root is None:
return
if len(ans) == depth:
ans.append(root.val)
dfs(root.right, depth + 1)
dfs(root.left, depth + 1)
ans = []
dfs(root, 0)
return ans(code-box)
/**
* 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 {
private List<Integer> ans = new ArrayList<>();
public List<Integer> rightSideView(TreeNode root) {
dfs(root, 0);
return ans;
}
private void dfs(TreeNode root, int depth) {
if (root == null) {
return;
}
if (ans.size() == depth) {
ans.add(root.val);
}
dfs(root.right, depth + 1);
dfs(root.left, depth + 1);
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> ans;
auto dfs = [&](this auto&& dfs, TreeNode* root, int depth) -> void {
if (!root) {
return;
}
if (ans.size() == depth) {
ans.push_back(root->val);
}
dfs(root->right, depth + 1);
dfs(root->left, depth + 1);
};
dfs(root, 0);
return ans;
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func rightSideView(root *TreeNode) (ans []int) {
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, depth int) {
if root == nil {
return
}
if len(ans) == depth {
ans = append(ans, root.Val)
}
dfs(root.Right, depth+1)
dfs(root.Left, depth+1)
}
dfs(root, 0)
return
}(code-box)
/**
* 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 = [];
const dfs = (root: TreeNode | null, depth: number) => {
if (!root) {
return;
}
if (ans.length == depth) {
ans.push(root.val);
}
dfs(root.right, depth + 1);
dfs(root.left, depth + 1);
};
dfs(root, 0);
return ans;
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var rightSideView = function (root) {
const ans = [];
const dfs = (root, depth) => {
if (!root) {
return;
}
if (ans.length == depth) {
ans.push(root.val);
}
dfs(root.right, depth + 1);
dfs(root.left, depth + 1);
};
dfs(root, 0);
return ans;
};(code-box)
// 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::rc::Rc;
impl Solution {
pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut ans = Vec::new();
fn dfs(node: Option<Rc<RefCell<TreeNode>>>, depth: usize, ans: &mut Vec<i32>) {
if let Some(node_ref) = node {
let node = node_ref.borrow();
if ans.len() == depth {
ans.push(node.val);
}
dfs(node.right.clone(), depth + 1, ans);
dfs(node.left.clone(), depth + 1, ans);
}
}
dfs(root, 0, &mut ans);
ans
}
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var rightSideView = function (root) {
const ans = [];
const dfs = (root, depth) => {
if (!root) {
return;
}
if (ans.length == depth) {
ans.push(root.val);
}
dfs(root.right, depth + 1);
dfs(root.left, depth + 1);
};
dfs(root, 0);
return ans;
};(code-box)
