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.

这道题就很简单了。反正就是递归，每遍历到一个点，就把前面计算到的数*10+当前数，到叶子结点的时候就把数加一下。

叶子结点就是左右结点都为NULL。

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int sumNumbers(TreeNode *root) {13         int sum = 0;   14         recursive(root, 0, sum);15         return sum;16     }17     18     void recursive(TreeNode* root, int value, int& sum) {19         if (root == NULL) {20             return;21         }22         int v = value * 10 + root->val;23         if (root->left == NULL && root->right == NULL) {24             sum += v;25         }26         recursive(root->left, v, sum);27         recursive(root->right, v, sum);28     }29 };

 第三遍刷，写了另外一种。

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int sumNumbers(TreeNode *root) {13         return recurse(root, 0);14     }15     16     int recurse(TreeNode *root, int sum) {17         if (root == NULL) return 0;18         if (root->left == NULL && root->right == NULL) return sum * 10 + root->val;19         int left = recurse(root->left, sum * 10 + root->val);20         int right = recurse(root->right, sum * 10 + root->val);21         return left + right;22     }23 };