Description#
Given the root of a binary tree, return an array of the largest value in each row of the tree (0-indexed).
Example 1:
Input: root = [1,3,2,5,3,null,9]
Output: [1,3,9]
Example 2:
Input: root = [1,2,3]
Output: [1,3]
Constraints:
- The number of nodes in the tree will be in the range
[0, 104]. -231 <= Node.val <= 231 - 1
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
| # 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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return []
q = deque([root])
ans = []
while q:
t = -inf
for _ in range(len(q)):
node = q.popleft()
t = max(t, node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(t)
return ans
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
| /**
* 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> largestValues(TreeNode root) {
List<Integer> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
int t = q.peek().val;
for (int i = q.size(); i > 0; --i) {
TreeNode node = q.poll();
t = Math.max(t, node.val);
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
ans.add(t);
}
return ans;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
| /**
* 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> largestValues(TreeNode* root) {
if (!root) return {};
queue<TreeNode*> q{{root}};
vector<int> ans;
while (!q.empty()) {
int t = q.front()->val;
for (int i = q.size(); i; --i) {
TreeNode* node = q.front();
t = max(t, node->val);
q.pop();
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
ans.push_back(t);
}
return ans;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func largestValues(root *TreeNode) []int {
var ans []int
if root == nil {
return ans
}
q := []*TreeNode{root}
for len(q) > 0 {
t := q[0].Val
for i := len(q); i > 0; i-- {
node := q[0]
q = q[1:]
t = max(t, node.Val)
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
ans = append(ans, t)
}
return ans
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
| /**
* 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 largestValues(root: TreeNode | null): number[] {
const res: number[] = [];
const queue: TreeNode[] = [];
if (root) {
queue.push(root);
}
while (queue.length) {
const n = queue.length;
let max = -Infinity;
for (let i = 0; i < n; i++) {
const { val, left, right } = queue.shift();
max = Math.max(max, val);
left && queue.push(left);
right && queue.push(right);
}
res.push(max);
}
return res;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
| // 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 largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = Vec::new();
let mut queue = VecDeque::new();
if root.is_some() {
queue.push_back(root.clone());
}
while !queue.is_empty() {
let mut max = i32::MIN;
for _ in 0..queue.len() {
let node = queue.pop_front().unwrap();
let node = node.as_ref().unwrap().borrow();
max = max.max(node.val);
if node.left.is_some() {
queue.push_back(node.left.clone());
}
if node.right.is_some() {
queue.push_back(node.right.clone());
}
}
res.push(max);
}
res
}
}
|
Solution 2#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| # 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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
def dfs(root, curr):
if root is None:
return
if curr == len(ans):
ans.append(root.val)
else:
ans[curr] = max(ans[curr], root.val)
dfs(root.left, curr + 1)
dfs(root.right, curr + 1)
ans = []
dfs(root, 0)
return ans
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
| /**
* 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> largestValues(TreeNode root) {
dfs(root, 0);
return ans;
}
private void dfs(TreeNode root, int curr) {
if (root == null) {
return;
}
if (curr == ans.size()) {
ans.add(root.val);
} else {
ans.set(curr, Math.max(ans.get(curr), root.val));
}
dfs(root.left, curr + 1);
dfs(root.right, curr + 1);
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
| /**
* 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> ans;
vector<int> largestValues(TreeNode* root) {
dfs(root, 0);
return ans;
}
void dfs(TreeNode* root, int curr) {
if (!root) return;
if (curr == ans.size())
ans.push_back(root->val);
else
ans[curr] = max(ans[curr], root->val);
dfs(root->left, curr + 1);
dfs(root->right, curr + 1);
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func largestValues(root *TreeNode) []int {
var ans []int
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, curr int) {
if root == nil {
return
}
if curr == len(ans) {
ans = append(ans, root.Val)
} else {
ans[curr] = max(ans[curr], root.Val)
}
dfs(root.Left, curr+1)
dfs(root.Right, curr+1)
}
dfs(root, 0)
return ans
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
| /**
* 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 largestValues(root: TreeNode | null): number[] {
const res = [];
const dfs = (root: TreeNode | null, depth: number) => {
if (root == null) {
return;
}
const { val, left, right } = root;
if (res.length == depth) {
res.push(val);
} else {
res[depth] = Math.max(res[depth], val);
}
dfs(left, depth + 1);
dfs(right, depth + 1);
};
dfs(root, 0);
return res;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
| // 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;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, depth: usize, res: &mut Vec<i32>) {
if root.is_none() {
return;
}
let node = root.as_ref().unwrap().borrow();
if res.len() == depth {
res.push(node.val);
} else {
res[depth] = res[depth].max(node.val);
}
Self::dfs(&node.left, depth + 1, res);
Self::dfs(&node.right, depth + 1, res);
}
pub fn largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = Vec::new();
Self::dfs(&root, 0, &mut res);
res
}
}
|
