NINJA GRANDPARENTS

Easy
0/40
Average time to solve is 15m
8 upvotes
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Problem statement

No one knows about Ninja, what he is thinking and what he wants to do. Now Ninja asks for the family tree from his father. His family tree is in the form of a binary tree and each node ( or we can say a family member ) is mentioned with some integer. Now his father asks him to write a code so that he is able to calculate the sum of all child nodes with even grandparents, that is, the grandparents node with an even integer value.

So your task is to calculate the sum of values of all nodes that have even valued grandparents. Grandparents refer to the parent of the parent node or we can say the two levels above node is a grandparent node.

Example :

Consider the following binary tree :

Example

So the sum of child nodes having even valued grandparents is ‘4 + 8 + 1 + 9 + 2 = 24’
Detailed explanation ( Input/output format, Notes, Images )

Input Format :

The first line contains an integer ‘T' which denotes the number of test cases to be run. Then the test cases follow.
The first line of each test case contains the tree elements 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 :

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

Output Format :

For each test case, print the sum of all child nodes having even valued grandparent nodes.

Print output of each test case in a separate line.

Note :

You are not required to print anything. It has already been taken care of. Just implement the function.

Constraints :

1 <= T <= 100
1 <= N <= 3000
0 <= data <= 10^9

Where ‘T’ is the total number of test cases, ‘N’ is the number of nodes of the tree, and ‘data’ is the integer value mentioned on the node.

Time Limit : 1 sec

Sample Input 1 :

2
22 3 8 4 8 1 9 -1 -1 -1 -1 -1 -1 -1 2 -1 -1
1 2 3 4 5 6 7 8 -1 -1 -1 -1 -1 -1 -1 -1 -1

Sample Output 1 :

24
8

Explanation for Sample 1:

For the first test case, the level order traversal of a given tree is 22 3 8 4 8 1 9 -1 -1 -1 -1 -1 -1 -1 2 -1 -1 . Thus, its first root is root number with value 22 and left, and right nodes of the 22 are nodes with 3 and 8 respectively and so on.
The binary tree according to the test case is:

Example

So the sum of child nodes of even valued grandparents is ‘4 + 8 + 1 + 9 + 2 = 24’  

For the second test case, the binary tree is :

Example

So the sum of child nodes is ‘8’ as only ‘8’ has ‘2’ as a grandparent.

Sample Input2:

2
6 7 8 2 7 1 3 9 -1 1 4 -1 -1 -1 5 -1 -1 -1 -1 -1 -1 -1 -1
-1

Sample Output2:

18
0
Hint

Can you traverse the whole tree?

Approaches (2)
Recursive Approach

So the idea here is to traverse the tree and for each node, we check if there is a grandparent that exists for that particular node or not. If it exists we check if it is even or not.

 

Algorithm:

 

  • So we use the recursive method for calculating the sum. For this, we declare three variables previous which store the previous node, grandparent which stores information about the grandparent node.
  • Now we start traversing our tree with the help of recursion first we call our recursion for the left subtree then for the right subtree.
  • And we update or current with node left or right depends on in which direction we are going and we update our previous with current and grandparent with previous.
  • Now we check if grandparent exists we further check if it is even we stored the sum of the current node else we recur further.
  • In this way, we arrive at our final sum when the recursive call ends or we can say our current becomes NULL.
Time Complexity

O(N), where ‘N’ is the total number of nodes in the tree.

 

As we are doing traversal of a tree and we need to visit O(N) nodes in the worst case.

Space Complexity

O(N), where ‘N’ is the total number of nodes in the tree.

 

Due to a recursion call stack, it will require O(N) extra space in the case of the bamboo tree (or one-sided tree).

Code Solution
(100% EXP penalty)
NINJA GRANDPARENTS
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