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:
     // 每个节点都充当单独的树来遍历一次
    int FindPath(TreeNode* root, int sum) {
        if (root == nullptr) return ans;
        dfs(root, sum);
        FindPath(root->left, sum);
        FindPath(root->right, sum);
        return ans;
    }

    // dfs 查询以某个节点为根的路径条数有多少
    void dfs(TreeNode* root, int sum) {
        if (root == nullptr) return;
        if (sum == root->val) ans++;
        dfs(root->left, sum - root->val);
        dfs(root->right, sum - root->val);
    }
private:
    int ans = 0;
};