Algorithm
Problem Name: 1161. Maximum Level Sum of a Binary Tree
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: root = [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.
Example 2:
Input: root = [989,null,10250,98693,-89388,null,null,null,-32127] Output: 2
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -105 <= Node.val <= 105
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int maxLevelSum(TreeNode root) {
if (root == null) {
return 0;
}
int currLevel = 1;
int maxSum = Integer.MIN_VALUE;
int maxSumLevel = 0;
Queue < TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
int currSum = 0;
while (size-- > 0) {
TreeNode removed = queue.remove();
currSum += removed.val;
if (removed.left != null) {
queue.add(removed.left);
}
if (removed.right != null) {
queue.add(removed.right);
}
}
if (maxSum < currSum) {
maxSum = currSum;
maxSumLevel = currLevel;
}
currLevel++;
}
return maxSumLevel;
}
}
Input
root = [1,7,0,7,-8,null,null]
Output
2
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxLevelSum = function(root) {
if (root == null) return 0
let res = 1
let cur = [root]
let next = []
let max = Number.MIN_SAFE_INTEGER
let sum = 0
let level = 1
while (cur.length) {
let node = cur.pop()
if (node.left) next.push(node.left)
if (node.right) next.push(node.right)
sum += node.val
if (cur.length === 0) {
cur = next
next = []
if (sum > max) {
res = level
max = sum
}
sum = 0
level++
}
}
return res
}
// DFS
const maxLevelSum = function(root) {
let result = {};
let recursion = function(root, level) {
if (result[level] !== undefined) {
result[level] += root.val;
} else {
result[level] = root.val;
}
if (root.left !== null) {
recursion(root.left, level + 1);
}
if (root.right !== null) {
recursion(root.right, level + 1);
}
};
recursion(root, 1);
let resultkey = 1;
let max = Number.MIN_VALUE;
for (let key of Object.keys(result)) {
if (result[key] > max) {
max = result[key];
resultkey = key;
}
}
return Number(resultkey);
};
Input
root = [1,7,0,7,-8,null,null]
Output
2
#3 Code Example with Python Programming
Code -
Python Programming
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def maxLevelSum(self, root: TreeNode) -> int:
levels, l, q = [], 1, [root]
while q:
levels.append([sum(node.val for node in q), l])
l += 1
q = [child for node in q for child in (node.left, node.right) if child]
return sorted(levels, key = lambda x: (x[0], -x[1]))[-1][1]Input
root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output
2
#4 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _1161_MaximumLevelSumOfABinaryTree
{
public int MaxLevelSum(TreeNode root)
{
var currentLevel = 1;
int maxSum = int.MinValue, maxLevel = 1;
var queue = new Queue < TreeNode>();
queue.Enqueue(root);
while (queue.Count > 0)
{
var sum = 0;
var size = queue.Count;
for (int i = 0; i < size; i++)
{
var node = queue.Dequeue();
sum += node.val;
if (node.left != null) queue.Enqueue(node.left);
if (node.right != null) queue.Enqueue(node.right);
}
if (sum > maxSum)
{
maxLevel = currentLevel;
maxSum = sum;
}
currentLevel++;
}
return maxLevel;
}
}
}
Input
root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output
2