Maximum Level Sum

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Average time to solve is 20m
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Problem statement

You are given a tree consisting of ‘N’ nodes. Your task is to find the level of the tree with the maximum sum.

The level of a node of the tree is the distance of the node from the root node + 1.

For example: 
You are given the tree:

altText

The Sum of all the levels are = [3, 9, 7], Here the level whose sum is maximum is level 2. Hence the answer is 2.
Example: 
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

Example

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. 

The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).
Note : 
The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of the input contains a single integer, 'T,’ denoting the number of test cases.

The first line of each test case contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example for further clarification.
Output Format: 
For each test case, print a single integer representing the level whose sum is maximum.

Print the output of each test case in a separate line.
Constraints: 
1 <= T <= 10
2 <= N <= 10^6
-10^8 <= nodeVal <= 10^8

Time Limit: 1 sec
Note: 
You do not need to print anything. It has already been taken care of. Just implement the function.
Sample Input 1:
2
3 4 5 2 -1 -1 5 -1 -1 -1 -1
1 2 3 4 -1 -1 5 6 7 -1 -1 -1 -1 -1 -1
Sample Output 1:
2
4
Explaination:
For the first test case:

altText

The Sum of all the levels are = [3, 9, 7], Here the level whose sum is maximum is level 2. Hence the answer is 2.

For the second test case:

alttext

The Sum of all levels is [1, 5, 9, 13], Here the level whose sum is maximum is level 4. Hence the answer is 4.
Sample input 2:
2
1 2 3 4 -1 -1 -1 -1 -1
10 20 11 -1 -1 -1 -1
Sample Output 2;
2    
2
Hint

Try using a depth-first search.

Approaches (2)
Depth-First Search

In this approach, we will use depth-first search, and at each call of the dfs we will keep track of the level and add the node value to the level, then we will have an array of level order sum of the whole tree. Then we can find the maximum sum from that level.

 

We create a function dfs(node, levelSum, level), here node is the current node, level is the level of the node, and levelSum is the array to store sums of each level in the tree.
 

Algorithm:

  • The dfs(node,levelSum, level) funtion
    • If node is null
      • Return from the function.
    • If size of levelSum is equal to level
      • Insert node.val into levelSum
    • Otherwise
      • Add node.val to levelSum[level]
    • Call dfs(node.left, levelSum, level + 1)
    • Call dfs(node.right, levelSum, level + 1)
  • Initialise an empty array levelSum
  • Call dfs(root, levelSum, 0)
  • Set maxLevel and maxSum as 0 and -infinity respectively.
  • iterate i through 0 to size of levelSum - 1
    • If levelSum[i] is greater than maxSum
      • Set maxSum as levelSum[i]
      • Set maxLevel as  i + 1
  • Return maxLevel.
Time Complexity

O(N), Where N is the number of nodes in the tree.

 

We are doing a depth-first search on the binary tree, which will cost O(N) time. Hence the overall time complexity is O(N).

Space Complexity

O(N), Where N is the number of nodes in the tree.

 

The space complexity of the recursion stack is O(N). Hence the overall space complexity is O(N).

Code Solution
(100% EXP penalty)
Maximum Level Sum
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