二叉树中和为某一值的路径(三)

public class Solution {
    private int res = 0;
    //dfs 以每个结点作为根查询路径
    public int findPath (TreeNode root, int sum) {
        //为空则返回
        if(root == null)
            return res;
        //查询以某结点为根的路径数
        helper(root, sum);
        //以其子结点为新根
        findPath(root.left, sum);
        findPath(root.right, sum);
        return res;
    }
    //dfs查询以某结点为根的路径数
    private void helper(TreeNode root, int sum){
        if(root == null)
            return;
        //符合目标值
        if(sum == root.val)
            res++;
        //进入子节点继续找
        helper(root.left, sum - root.val);
        helper(root.right, sum - root.val);
    }
}