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.

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.

题意：路径组成数字，计算这些路径的和。

思路：找出各个路径的数字，然后求和。

/** * 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 {public:int sumNumbers(TreeNode* root) {if (root == NULL)return 0;int sum = 0;getPathString(root, "", sum);return sum;}private:void getPathString(TreeNode* root, string cur, int& sum){if (root->left == NULL  && root->right == NULL){cur += to_string(root->val);sum += atoi(cur.c_str());return;}if (root->left){getPathString(root->left, cur + to_string(root->val), sum);}if (root->right){getPathString(root->right, cur + to_string(root->val), sum);}return;}};

思路2：上述思路路径信息的保存用的是string，以下思路路径信息用int值存储。

/** * 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 {public:int sumNumbers(TreeNode* root) {if (root == NULL)return 0;int sum = 0;getPathString(root, 0, sum);return sum;}private:void getPathString(TreeNode* root, int cur, int& sum){cur *= 10;cur += root->val;if (root->left == NULL  && root->right == NULL){sum += cur;return;}if (root->left){getPathString(root->left, cur, sum);}if (root->right){getPathString(root->right, cur, sum);}return;}};