题解 | #二叉树中和为某一值的路径(三)#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型
     */
    int result = 0;
    map<int, int> mp;

    void rec(TreeNode* root, int now_sum, int target)
    {
        if(root==NULL)
            return;
        int now = now_sum+root->val;
      
        if(now==target)
        {
            result+=1;
        }

        result+=mp[now-target];
 
        mp[now]++;
        rec(root->left,now,target);
        rec(root->right,now,target);
        mp[now]--;
    }


    int FindPath(TreeNode* root, int sum) {
        // write code here
        if(root==NULL)
            return 0;
        rec(root,0,sum);
        return result;
    }
};