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 

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. 

Implementation:

C++

// C++ program to find sum of
// all paths from root to leaves
#include <bits/stdc++.h>
using namespace std;

class node
{
    public:
    int data;
    node *left, *right;
};

// function to allocate new node with given data
node* newNode(int data)
{
    node* Node = new 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(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(node *root)
{
    // Pass the initial value as 0
    // as there is nothing above root
    return treePathsSumUtil(root, 0);
}

// Driver code
int main()
{
    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);
    cout<<"Sum of all paths is "<<treePathsSum(root);
    return 0;
}

// This code is contributed by rathbhupendra

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 %d", 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

Python3

# 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

# Returns sums of all root to leaf paths. The first parameter is root
# of current subtree, the second paramete"r 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)

C#

// c# program to find sum of all numbers
// that are formed from root to leaf paths
using System;

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

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

class GFG
{
public 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
public virtual 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()
public virtual int treePathsSum(Node node)
{
    // Pass the initial value as 0 as
    // there is nothing above root
    return treePathsSumUtil(node, 0);
}

// Driver Code
public static void Main(string[] args)
{
    GFG tree = new GFG();
    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);

    Console.Write("Sum of all paths is " +
                   tree.treePathsSum(tree.root));
}
}

// This code is contributed by Shrikant13

Javascript

<script>
// JavaScript program to find sum of
// all paths from root to leaves
class node
{

    constructor(data){
        this.data = data;
        this.left = this.right = null;
    }

}

// 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
function treePathsSumUtil(root,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()
function treePathsSum(root)
{
    // Pass the initial value as 0
    // as there is nothing above root
    return treePathsSumUtil(root, 0);
}

// Driver code

let root = new node(6);
root.left = new node(3);
root.right = new node(5);
root.left.left = new node(2);
root.left.right = new node(5);
root.right.right = new node(4);
root.left.right.left = new node(7);
root.left.right.right = new node(4);
document.write("Sum of all paths is "+treePathsSum(root));

// This code is contributed by shinjanpatra
</script>

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.
Auxiliary Space: O(n)

Another Approach: We can also solve this problem by first finding all the paths from root to leaf .Then we convert all paths in to the numbers .In the end we will add those numbers.

Implementation:

Python3

# 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


def treePathsSumUtil(root, currPath, allPath):

    # Base Case
    if root is None:
        return

    # append the root data in string format in currPath
    currPath.append(str(root.data))

    # if we found a leaf node we copy the the currPath to allPath
    if root.left is None and root.right is None:
        allPath.append(currPath.copy())

    # traverse in the left subtree
    treePathsSumUtil(root.left, currPath, allPath)

    # traverse in the right subtree
    treePathsSumUtil(root.right, currPath, allPath)

    # remove the current element from the path
    del currPath[-1]


def treePathsSum(root):
    # store all the root to leaf path in allPath
    allPath = []

    treePathsSumUtil(root, [], allPath)
    # store the sum
    s = 0

    for pathNumber in allPath:
        # join the pathNumbers to convert them  into the number to calculate sum
        k = "".join(pathNumber)
        s += int(k)
    return s


# 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 Vivek Maddeshiya

Output

Sum of all paths is 13997

Time Complexity:- Time Complexity of this approach will be O(n^2)  because we are traversing the allPath and joining currPath to  the allPath array .
Auxiliary Space:-  O(n)