# 129. 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 1:

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.

# Solution

Approach 1: recursive.

# Code (Python)

Approach 1:

    def sumNumbers(self, root: 'TreeNode') -> 'int':
        return self._sum(root, 0)
    
    def _sum(self, node, sum_so_far):
        if not node:
            return 0
        if not node.left and not node.right:
            return sum_so_far * 10 + node.val
        new_sum = 0
        if node.left:
            new_sum += self._sum(node.left, sum_so_far * 10 + node.val)
        if node.right:
            new_sum += self._sum(node.right, sum_so_far * 10 + node.val)
        return new_sum

# Code (C++)

Approach 1:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    int totalSum = 0;
    void doSumNumbers(TreeNode* root, int sum) {
        if (root == NULL)
            return;
        sum = sum * 10 + root->val;
        if (!root->left && !root->right)
        {
            totalSum += sum;
            return;
        }
        if (root->left)
            doSumNumbers(root->left, sum);
        if (root->right)
            doSumNumbers(root->right, sum);
    }
public:
    int sumNumbers(TreeNode* root) {
        doSumNumbers(root, 0);
        return totalSum;
    }
};