Ninja went to a school as an invigilator for an exam the seats in the room were mentioned with the number and they form a tree-like structure so he started thinking of a solution so that he allocated seats to students in such a way that no students would sit adjacent to each other.
So help our ninja write a code for the problem so he is able to count the maximum number of seats available for the students that no student can sit adjacent to each other.
So your task is to write a code that can find the maximum subset of a node in a given binary tree where there is no edge between any two nodes of the subset.
For example, consider the following binary tree. The largest subset set is {1, 5, 6, 7} and the size of the subset is 4.
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the 'T' test cases are as follow.
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 below.
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 :

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 -1 5 6 7 -1 -1 -1 -1
Output Format :
For each test case, print the maximum number of elements in a subset satisfying the given condition. Print output of each test case in a separate line.
Note:
You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
1 <= T <= 100
1 <= N <= 3000
-10^9 <= DATA <= 10^9
Where ‘T’ is the total number of test cases, ‘N’ is the number of benches(nodes of the tree) in the given class, and 'DATA' is the value mentioned on the bench(value of nodes).
Time Limit: 1sec
2
1 2 3 -1 4 -1 5 -1 -1 -1 -1
1 2 -1 4 -1 -1 -1
3
2
In first test case, the level order traversal of a given class is 1 2 3 -1 4 -1 5 -1 -1 -1 -1, thus its first bench is bench number with value 1, and left and right benches of the 1 are nodes with 2 and 3 respectively.
The binary tree according to the test case is:
So we choose the bench with the bench number as {1, 4, 5} so the output is 3.
In the second test case, the binary tree according to the test case is:
So we choose the bench with the bench number as {1, 4} so the output is 2.
2
1 -1 2 -1 -1
1 2 3 4 5 -1 6 -1 -1 7 8 -1 -1 -1 -1 -1 -1
1
5
In the first test case, there are only two nodes thus, one of them is only useful and so, answer will be 1.
In the second test case, {1, 4, 6 , 7, 8} can be used and so, answer will be 5.
Can we find the largest subset for a node X if we know the largest subset for all descendants of X?
Approach: We use recursion by using simple two possibilities for every node x, either X is a member of the set or not a member:
Here is the algorithm:
O(2^N), where N denotes the number of nodes in the binary tree.
We are calling 2 child calls ( either members or not) at every stage. So for all ‘N’ calls total will require ‘2^N’.
O(N), where ‘N’ denotes the number of nodes in a binary tree.
We are using a call stack of size ‘N’ due to recursion.