/*

 sumUp递归子程序求解以root为根节点的子节点之和为sum的路径数目;

 pathSum递归部分是把根节点逐一考察，如以root->left，以root->right为根等等，

 用sumUp求解其对应的路径数目。最终结果是全部求和，也就是pathSum return的形式。

 */

 class Solution {

 public:

     int pathSum(TreeNode* root, int sum)

     {

         if(!root) return ;

         return sumUp(root, , sum) + pathSum(root->left, sum) + pathSum(root->right, sum);

     }

 private:

     int sumUp(TreeNode* root, int pre, int& sum)

     {

         if(!root) return ;

         int current = pre + root->val;

         return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);

     }

 }