给定一个二叉树的 根节点 root，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。
示例 1:
输入: [1,2,3,null,5,null,4] 输出: [1,3,4]
示例 2:
输入: [1,null,3] 输出: [1,3]
示例 3:
输入: [] 输出: []
提示:
- 二叉树的节点个数的范围是
[0,100] -100 <= Node.val <= 100
使用 BFS 层序遍历二叉树，每层最后一个节点即为该层的右视图节点。
时间复杂度
# 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[-1].val)
for _ in range(len(q)):
node = q.popleft()
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return ans/**
* 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.peekLast().val);
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return ans;
}
}/**
* 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.empty()) {
ans.emplace_back(q.back()->val);
for (int n = q.size(); n; --n) {
TreeNode* node = q.front();
q.pop();
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
}
return ans;
}
};/**
* 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[len(q)-1].Val)
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
return
}/**
* 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 = [];
if (!root) {
return ans;
}
const q = [root];
while (q.length) {
const n = q.length;
ans.push(q[n - 1].val);
for (let i = 0; i < n; ++i) {
const { left, right } = q.shift();
left && q.push(left);
right && q.push(right);
}
}
return ans;
}// 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::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
if root.is_none() {
return res;
}
let mut q = VecDeque::new();
q.push_back(root);
while !q.is_empty() {
let n = q.len();
res.push(q[n - 1].as_ref().unwrap().borrow().val);
for _ in 0..n {
if let Some(node) = q.pop_front().unwrap() {
let mut node = node.borrow_mut();
if node.left.is_some() {
q.push_back(node.left.take());
}
if node.right.is_some() {
q.push_back(node.right.take());
}
}
}
}
res
}
}使用 DFS 深度优先遍历二叉树，每次先遍历右子树，再遍历左子树，这样每层第一个遍历到的节点即为该层的右视图节点。
时间复杂度
# 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(node, depth):
if node is None:
return
if depth == len(ans):
ans.append(node.val)
dfs(node.right, depth + 1)
dfs(node.left, depth + 1)
ans = []
dfs(root, 0)
return ans/**
* 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 node, int depth) {
if (node == null) {
return;
}
if (depth == ans.size()) {
ans.add(node.val);
}
dfs(node.right, depth + 1);
dfs(node.left, depth + 1);
}
}/**
* 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;
function<void(TreeNode*, int)> dfs = [&](TreeNode* node, int depth) {
if (!node) {
return;
}
if (depth == ans.size()) {
ans.emplace_back(node->val);
}
dfs(node->right, depth + 1);
dfs(node->left, depth + 1);
};
dfs(root, 0);
return ans;
}
};/**
* 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(node *TreeNode, depth int) {
if node == nil {
return
}
if depth == len(ans) {
ans = append(ans, node.Val)
}
dfs(node.Right, depth+1)
dfs(node.Left, depth+1)
}
dfs(root, 0)
return
}/**
* 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 = (node: TreeNode | null, depth: number) => {
if (!node) {
return;
}
if (depth == ans.length) {
ans.push(node.val);
}
dfs(node.right, depth + 1);
dfs(node.left, depth + 1);
};
dfs(root, 0);
return ans;
}