##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1161.html

# 1161. Maximum Level Sum of a Binary Tree (Medium)

Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.

Return the smallest level X such that the sum of all the values of nodes at level X is maximal.

Example 1:

Input: [1,7,0,7,-8,null,null]
Output: 2
Explanation:
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.


Note:

1. The number of nodes in the given tree is between 1 and 10^4.
2. -10^5 <= node.val <= 10^5

Companies:

Related Topics:
Graph

## Solution 1. Level-order Traversal

// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxLevelSum(TreeNode* root) {
if (!root) return 0;
int maxSum = 0, ans = 0, lv = 1;
queue<TreeNode*> q;
q.push(root);
while (q.size()) {
int cnt = q.size(), sum = 0;
while (cnt--) {
auto node = q.front();
q.pop();
sum += node->val;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
if (sum > maxSum) {
maxSum = sum;
ans = lv;
}
++lv;
}
return ans;
}
};

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxLevelSum(TreeNode root) {
if (root == null)
return 0;
int maxSum = 0;
int maxSumLevel = 0;
int level = 0;
queue.offer(root);
while (!queue.isEmpty()) {
level++;
int curSum = 0;
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
curSum += node.val;
TreeNode left = node.left, right = node.right;
if (left != null)
queue.offer(left);
if (right != null)
queue.offer(right);
}
if (curSum > maxSum) {
maxSum = curSum;
maxSumLevel = level;
}
}
return maxSumLevel;
}
}

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

/**
* 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 int maxLevelSum(TreeNode root) {
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int mx = Integer.MIN_VALUE;
int i = 0;
int ans = 0;
while (!q.isEmpty()) {
++i;
int s = 0;
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.pollFirst();
s += node.val;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
if (mx < s) {
mx = s;
ans = i;
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxLevelSum(TreeNode* root) {
if (!root) return 0;
int maxSum = 0, ans = 0, lv = 1;
queue<TreeNode*> q;
q.push(root);
while (q.size()) {
int cnt = q.size(), sum = 0;
while (cnt--) {
auto node = q.front();
q.pop();
sum += node->val;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
if (sum > maxSum) {
maxSum = sum;
ans = lv;
}
++lv;
}
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 maxLevelSum(self, root: Optional[TreeNode]) -> int:
q = deque([root])
mx = -inf
i = 0
while q:
i += 1
s = 0
for _ in range(len(q)):
node = q.popleft()
s += node.val
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if mx < s:
mx = s
ans = i
return ans

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

# 1161. Maximum Level Sum of a Binary Tree
# https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/

# 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 maxLevelSum(self, root: TreeNode) -> int:
level = res = 1
mmax = float('-inf')
q = collections.deque([root])

while q:
val = 0
n = len(q)

for _ in range(n):
node = q.popleft()
val += node.val
for leaf in (node.left, node.right):
if leaf:
q.append(leaf)

if val > mmax:
res = level
mmax = val

level += 1

return res


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func maxLevelSum(root *TreeNode) int {
q := []*TreeNode{root}
mx := -0x3f3f3f3f
i := 0
ans := 0
for len(q) > 0 {
i++
s := 0
for n := len(q); n > 0; n-- {
root = q[0]
q = q[1:]
s += root.Val
if root.Left != nil {
q = append(q, root.Left)
}
if root.Right != nil {
q = append(q, root.Right)
}
}
if mx < s {
mx = s
ans = i
}
}
return ans
}

• /**
* 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 maxLevelSum(root: TreeNode | null): number {
const queue = [root];
let res = 1;
let max = -Infinity;
let h = 1;
while (queue.length !== 0) {
const n = queue.length;
let sum = 0;
for (let i = 0; i < n; i++) {
const { val, left, right } = queue.shift();
sum += val;
left && queue.push(left);
right && queue.push(right);
}
if (sum > max) {
max = sum;
res = h;
}
h++;
}
return res;
}