257. Binary Tree Paths (Easy)

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

Solution: DFS

这道题给我们一个二叉树，让我们返回所有根到叶节点的路径，跟之前那道Path Sum II 二叉树路径之和之二很类似，比那道稍微简单一些，不需要计算路径和，只需要无脑返回所有的路径即可，那么思路还是用DFS来解，代码而很简洁，参见如下：

/**
 * 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 {
    void dfs(TreeNode* root, string path, vector<string> &res) {
        path += to_string(root->val);
        if (!root->left && !root->right) res.push_back(path);
        if (root->left) dfs(root->left, path+"->", res);
        if (root->right) dfs(root->right, path+"->", res);
    }
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root) dfs(root, "", res);
        return res;
    }
};