Odd even level

Easy
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Average time to solve is 20m
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ZSAmazonCitrix

Problem statement

Given a binary tree. Find and return the modulus of the difference between the sum of odd level nodes and the sum of even level nodes.

Detailed explanation ( Input/output format, Notes, Images )
Input format: 
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then, the test cases follow.

The first and only line of every test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 on its place.


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

alt text

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
Output format: 
For each test case, print an integer denoting the modulus of the difference in a single line.
Constraints: 
1 <= T <= 5
0 <= N <= 10^5
0 <= val <= 10^9

Time limit: 1 sec
Sample Input 1:
1
5 6 10 2 3 5 9 -1 -1 -1 -1 -1 -1 -1 -1
Sample Output 1:
8

Explanation Binary tree

Level 1= {5}, Sum= 5
Level 2= {6, 10} Sum= 16
Level 3= {2, 3, 5, 9}, Sum=19
Odd sum = Sum of Level 1 + Sum of Level 3
                   5 + 19 
                   24

Even sum = Sum of Level 2
                    16

| Even sum - Odd sum | = 8
Sample Input 2:
1
5 3 9 2 6 5 10 -1 -1 -1 -1 -1 -1 -1 -1
Sample Output 2:
16
Hint

Although the question is focusing on levels of the tree, the use of queue can be eliminated by using recursion i.e depth-first search traversal of the tree. We can maintain the level we are traversing on by using an extra parameter in the DFS function.    

Approaches (2)
DFS
  1. Create two variables, one for storing the sum of nodes at the odd level and one for storing the sum of nodes at even level.
  2. Use the below algorithm to calculate level sums where:

oddSum: denotes sum of nodes at odd levels, initially 0.

evenSum: denotes sum of nodes at ecen levels, initially 0.

level: the level of the current node, initially 1. 

void oddEvenLevelHelper(current, oddSum, evenSum, level)

 

  • If the current level is odd, update the oddSum.
  • Otherwise, update evenSum.
  • Recursively call for the left subtree with the updated level.
  • Recursively call for the right subtree with the updated level.

 

Return the modulus of the difference between oddSum and evenSum.

Time Complexity

O(N) where N is the number of tree nodes. 

In the worst case, we will be traversing the whole binary tree once.   

Space Complexity

Space complexity

O(L) where L is the depth of the binary tree.

In the worst case, O(N) if it is a skew tree where N is the number of nodes in the Binary tree. 

We are performing recursion which uses stack space and the stack fills maximum up to the depth of the binary tree.  

Code Solution
(100% EXP penalty)
Odd even level
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