# Sum of all the numbers that are formed from root to leaf paths

Given a binary tree, where every node value is a Digit from 1-9 .Find the sum of all the numbers which are formed from root to leaf paths.

For example consider the following Binary Tree.

```                                          6
/      \
3          5
/   \          \
2     5          4
/   \
7     4
There are 4 leaves, hence 4 root to leaf paths:
Path                    Number
6->3->2                   632
6->3->5->7               6357
6->3->5->4               6354
6->5>4                    654
Answer = 632 + 6357 + 6354 + 654 = 13997 ```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to do a preorder traversal of the tree. In the preorder traversal, keep track of the value calculated till the current node, let this value be val. For every node, we update the val as val*10 plus node’s data.

## C

```// C program to find sum of all paths from root to leaves
#include <stdio.h>
#include <stdlib.h>

struct node
{
int data;
struct node *left, *right;
};

// function to allocate new node with given data
struct node* newNode(int data)
{
struct node* node = (struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = node->right = NULL;
return (node);
}

// Returns sum of all root to leaf paths. The first parameter is root
// of current subtree, the second parameter is value of the number formed
// by nodes from root to this node
int treePathsSumUtil(struct node *root, int val)
{
// Base case
if (root == NULL)  return 0;

// Update val
val = (val*10 + root->data);

// if current node is leaf, return the current value of val
if (root->left==NULL && root->right==NULL)
return val;

// recur sum of values for left and right subtree
return treePathsSumUtil(root->left, val) +
treePathsSumUtil(root->right, val);
}

// A wrapper function over treePathsSumUtil()
int treePathsSum(struct node *root)
{
// Pass the initial value as 0 as there is nothing above root
return treePathsSumUtil(root, 0);
}

// Driver function to test the above functions
int main()
{
struct node *root = newNode(6);
root->left        = newNode(3);
root->right       = newNode(5);
root->left->left  = newNode(2);
root->left->right = newNode(5);
root->right->right = newNode(4);
root->left->right->left = newNode(7);
root->left->right->right = newNode(4);
printf("Sum of all paths is", treePathsSum(root));
return 0;
}
```

## Java

```// Java program to find sum of all numbers that are formed from root
// to leaf paths

// A binary tree node
class Node
{
int data;
Node left, right;

Node(int item)
{
data = item;
left = right = null;
}
}

class BinaryTree
{
Node root;

// Returns sum of all root to leaf paths. The first parameter is
// root of current subtree, the second parameter is value of the
// number formed by nodes from root to this node
int treePathsSumUtil(Node node, int val)
{
// Base case
if (node == null)
return 0;

// Update val
val = (val * 10 + node.data);

// if current node is leaf, return the current value of val
if (node.left == null && node.right == null)
return val;

// recur sum of values for left and right subtree
return treePathsSumUtil(node.left, val)
+ treePathsSumUtil(node.right, val);
}

// A wrapper function over treePathsSumUtil()
int treePathsSum(Node node)
{
// Pass the initial value as 0 as there is nothing above root
return treePathsSumUtil(node, 0);
}

// Driver program to test above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(6);
tree.root.left = new Node(3);
tree.root.right = new Node(5);
tree.root.right.right = new Node(4);
tree.root.left.left = new Node(2);
tree.root.left.right = new Node(5);
tree.root.left.right.right = new Node(4);
tree.root.left.right.left = new Node(7);

System.out.print("Sum of all paths is " +
tree.treePathsSum(tree.root));
}
}

// This code has been contributed by Mayank Jaiswal

```

## Python

```
# Python program to find sum of all paths from root to leaves

# A Binary tree node
class Node:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

# Returs sums of all root to leaf paths. The first parameter is root
# of current subtree, the second parameter is value of the number
# formed by nodes from root to this node
def treePathsSumUtil(root, val):

# Base Case
if root is None:
return 0

# Update val
val = (val*10 + root.data)

# If current node is leaf, return the current value of val
if root.left is None and root.right is None:
return val

# Recur sum of values for left and right subtree
return (treePathsSumUtil(root.left, val) +
treePathsSumUtil(root.right, val))

# A wrapper function over treePathSumUtil()
def treePathsSum(root):

# Pass the initial value as 0 as ther is nothing above root
return treePathsSumUtil(root, 0)

# Driver function to test above function
root = Node(6)
root.left = Node(3)
root.right = Node(5)
root.left.left = Node(2)
root.left.right = Node(5)
root.right.right = Node(4)
root.left.right.left = Node(7)
root.left.right.right = Node(4)
print "Sum of all paths is", treePathsSum(root)

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

```

Output:
`Sum of all paths is 13997`

Time Complexity: The above code is a simple preorder traversal code which visits every exactly once. Therefore, the time complexity is O(n) where n is the number of nodes in the given binary tree.

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