root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
​
      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1
​
返回 3。和等于 8 的路径有:
​
1.  5 -> 3
2.  5 -> 2 -> 1
3.  -3 -> 11

/**
 * 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 res=0;
    void getsum(TreeNode* root, int sum)
    {
        if(!root) return ;
        sum-=root->val;
        if(sum==0) res++;
        getsum(root->left,sum);
        getsum(root->right,sum);
    }
public:
    int pathSum(TreeNode* root, int sum) {
​
        if(!root) return 0;
        getsum(root,sum);
        pathSum(root->left,sum);
        pathSum(root->right,sum);
        return res;
​
​
​
​
    }
};