# 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  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: #include 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 "<

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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