Last Updated: 12 Oct, 2020
Odd even level
Easy
Asked in companies
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.
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 :
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
Approaches
- 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.
- 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.
- Create a queue of Binary tree nodes.
- Initialise two variables, one for storing the sum of odd level nodes and one for storing the sum of even level nodes.
- Keep on updating the queue level by level while maintaining the sum of odd level nodes as well as the sum of even level nodes.
- Return the modulus of the difference between the sum of odd level nodes and the sum of even level nodes.
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