# Sum root to leaf

Posted: 17 Jan, 2021

Difficulty: Easy

#### You are given an arbitrary binary tree consisting of N nodes where each node is associated with a certain integer value from 1 to 9. Consider each root to leaf path as a number.

#### For example:

```
1
/ \
2 3
The root to leaf path 1->2 represents the number 12.
The root to leaf path 1->3 represents the number 13.
```

#### Your task is to find the total sum of all the possible root to leaf paths.

#### In the above example,

```
The total sum of all the possible root to leaf paths is 12+13 = 25
```

##### Note:

```
The output may be very large, return the answer after taking modulus with (10^9+7).
```

##### Input format

```
The first line of input contains an integer ‘T’ denoting the number of test cases.
The next ‘T’ lines represent the ‘T’ test cases.
The only 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 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
```

##### Output format:

```
For each test case, return the total sum of all the possible root to leaf paths.
Output for each test case must be in a separate line.
```

##### Note :

```
You do not need to print anything, it has already been taken care of. Just implement the given function.
```

##### Constraints:

```
1 <= T <= 50
1 <= N <= 3000
1 <= nodeValue <= 9
Where ‘N’ is the number of nodes in the tree and nodeValue denotes data contained in the node of a binary tree.
Time limit: 1 sec
```

Approach 1

The idea is to traverse the tree using preorder traversal and keep calculating the number formed till the current node. Once we reach the leaf node, add the number obtained to the total sum.

- If the ‘ROOT’ is ‘NULL’, return 0.
- For each node, form the number till the current node and store it in the variable ‘VALUE’.
- If the node is a leaf node, return the number formed.
- Otherwise, recur with the addition of values obtained from the right and left subtrees.
- Make sure to use modulo operation while returning value from every function call.

SIMILAR PROBLEMS

# Balanced Binary Tree

Posted: 1 Jul, 2021

Difficulty: Moderate

# Postorder Traversal

Posted: 22 Jul, 2021

Difficulty: Easy

# Preorder Traversal

Posted: 22 Jul, 2021

Difficulty: Easy

# Maximum Sum BST

Posted: 27 Jul, 2021

Difficulty: Hard

# Vertical Sum in BST

Posted: 27 Jul, 2021

Difficulty: Moderate