二叉树根节点到叶子节点路径和为指定值的所有路径

先序遍历，套模版就是了，没啥可解释的。

class Solution {
public:
    /**
     *
     * @param root TreeNode类
     * @param sum int整型
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > pathSum(TreeNode* root, int sum) {
        // write code here
        vector<vector<int>> res;
        if (!root) return res;
        vector<int> path;
        preOrder(root, res, path, sum, 0);
        return res;
    }

    void preOrder(TreeNode* root, vector<vector<int> > &res, vector<int> path, int sum, int current) {
        if (!root) return;
        current = current + root->val;
        path.push_back(root->val);
        if (!root->left && !root->right && sum == current) res.push_back(path);
        if (root->left) preOrder(root->left, res, path, sum ,current);
        if (root->right) preOrder(root->right, res, path, sum ,current);
    }
};