257. Binary Tree Paths

做题历程：

1. 本题做了不止2次了，本次大概耗时7分钟，独立解出

本题难度还是比较低的，就是单纯的用DFS就能解出，就不赘述了，直接上代码吧：

/**
 * 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 {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        if (root != nullptr) {
            dfs(root, "");
        }
        return res;
    }
private:
    vector<string> res;
    void dfs(TreeNode* root, string s) {
        if (s == "") {
            s = s + to_string(root->val);
        }
        else {
            s = s + "->" + to_string(root->val);
        }

        if (root->left == nullptr && root->right == nullptr) {
            res.push_back(s);
            return;
        }
        if (root->left != nullptr) {
            dfs(root->left, s);
        }
        if (root->right != nullptr) {
            dfs(root->right, s);
        }
    }
};