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.
For example,
1 / \ 2 3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Return the sum = 12 + 13 = 25.
这道求根到叶节点数字之和的题跟之前的求Path Sum 二叉树的路径和很类似，都是利用DFS递归来解，这道题由于不是单纯的把各个节点的数字相加，而是每到一个新的数字，要把原来的数字扩大10倍之后再相加。代码如下：
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int sumNumbers(TreeNode *root) { return sumNumbersDFS(root, 0); } int sumNumbersDFS(TreeNode *root, int sum) { if (!root) return 0; sum = sum * 10 + root->val; if (!root->left && !root->right) return sum; return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right, sum); } };
本文转自博客园Grandyang的博客，原文链接：求根到叶节点数字之和[LeetCode] Sum Root to Leaf Numbers ，如需转载请自行联系原博主。
