# NINJA GRANDPARENTS

Posted: 27 Mar, 2021

Difficulty: Easy

#### 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 :
```

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

#### 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 :
```

```
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
```

Approach 1

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.

Approach 2

So the idea here is to traverse the tree by using bfs or we can say with the help of queue 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.

- So we start our traversal by declaring a queue and first we push our root node into it.
- Now we run a loop that will run till our queue becomes empty and if the left node of the tree exists we push that into our queue and do the same for the right node.
- Now we check if the data of some node is even and add its grandchild sum into our ‘ans’ variable.
- We check whether the node has a left grandchild or a right grandchild or both accordingly we add that into our answer.
- Now we can simply return our ‘ans’ variable.

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