# 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;
}
};