# Daily Coding Challenge #45

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

``````/*
Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
1
/ \
2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:

Input: [4,9,0,5,1]
4
/ \
9   0
/ \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
*/

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int solve(TreeNode *root, int sum){
// recursive approach - add *10 for each layer and plus its value
if(!root) return 0;
if(!root->left&&!root->right) return sum*10+root->val;
return solve(root->left, sum*10+root->val)+solve(root->right, sum*10+root->val);
}
int sumNumbers(TreeNode* root) {
return solve(root,0);
}
};
``````

The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### Discussion  stjrna

hello ;)
if it were an array you can do it like this .. oh yes i didnt compile so figure to fix code .. remember nothing complete.. happy coding

int A[] = {4,9,0,5,1};
int len = sizeof(A)/sizeof(A*); // length
int i=0; int L=0;

while(i<len-3)
{
L = ((i+1)*2)-1;

if (L<len-3) A[L++]+=A[i]*10;
if (L<len-3) A[L]+=A[i++]*10;
}
L=0;
while(len-i) L+=A[i]+A[i++ +1];  