##### Welcome to Subscribe On Youtube
• import java.util.ArrayList;
import java.util.List;
import java.util.Queue;

/**

You need to find the largest value in each row of a binary tree.

Example:
Input:

1
/ \
3   2
/ \   \
5   3   9

Output: [1, 3, 9]

*/

public class Find_Largest_Value_in_Each_Tree_Row {

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> largestValues(TreeNode root) {

List<Integer> result = new ArrayList<>();

// bfs
if (root == null) {
return result;
}

q.offer(root);
int currentLevelCount = 1;
int nextLevelCount = 0;

int currentLevelMax = root.val;

while (!q.isEmpty()) {
TreeNode current = q.poll();
currentLevelCount--;

// check leftmost
currentLevelMax = Math.max(currentLevelMax, current.val);

if (current.left != null) {
q.offer(current.left);
nextLevelCount++;
}
if (current.right != null) {
q.offer(current.right);
nextLevelCount++;
}

if (currentLevelCount == 0) {
currentLevelCount = nextLevelCount;
nextLevelCount = 0;

// reset
currentLevelMax = Integer.MIN_VALUE;
}
}

return result;
}
}
}

############

/**
* 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);
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/find-largest-value-in-each-tree-row/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> largestValues(TreeNode* root) {
if (!root) return {};
vector<int> ans;
queue<TreeNode*> q{ {root} };
while (q.size()) {
int cnt = q.size(), maxVal = INT_MIN;
while (cnt--) {
auto n = q.front();
q.pop();
maxVal = max(maxVal, n->val);
if (n->left) q.push(n->left);
if (n->right) q.push(n->right);
}
ans.push_back(maxVal);
}
return ans;
}
};

• # 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

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def largestValues(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
ans = []
d = {}

def dfs(root, h, d):
if root:
dfs(root.left, h + 1, d)
dfs(root.right, h + 1, d)
d[h] = max(d.get(h, float("-inf")), root.val)

dfs(root, 0, d)
level = 0
while level in d:
ans += d[level],
level += 1
return ans


• /**
* 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• /**
* 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;
}


• // 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
}
}