257.Binary Tree Paths

给定一个二叉树，返回所有根节点到叶子节点的路径。
Input:
 1
 /
2  3

5
Output: ["1->2->5", "1->3"]
Explanation: All root-to-leaf paths are: 1->2->5, 1->3

难点：返回的是vector<string>的容器类型，需要对容器遍历读写操作。
思路：运用3个容器，分别为接收左子树路径的 res_left，接收右子树路径的 res_right，和总的路径的 res. 对每次遍历所得到的 res_left 和 res_right 遍历，将里面的已有路径读出来，加上当前节点的值，再写入到总的路径容器 res 中，最后返回 res.

vector<string> binaryTreePaths(TreeNode* root) {
    if (!root) return {};
    vector<string> res_left;
    vector<string> res_right;
    vector<string> res;
    if (root->left != NULL) res_left = binaryTreePaths(root->left);
    if (root->right != NULL) res_right = binaryTreePaths(root->right);
    if (res_left.size() == 0 && res_right.size() == 0) {
        return{ to_string(root->val) };
    }
    for (int i = 0; i < res_left.size(); i++) {
        res.push_back(to_string(root->val) + "->" + res_left[i]);
    }
    for (int i = 0; i < res_right.size(); i++) {
        res.push_back(to_string(root->val) + "->" + res_right[i]);
    }
    return res;
}

Java版：

刷第二遍，感觉利用递归 + 深搜，将每一次遍历的路线都记录下来，当节点是叶子节点时，将记录下来的路径加到结果集中。

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> ans = new ArrayList<>();
        binaryTreePathsDFS(root, "", ans);
        return ans;
    }

    public void binaryTreePathsDFS(TreeNode root, String path, List<String> ans){
        if(root == null) return;
        path += root.val;
        if(root.left == null && root.right == null) ans.add(path);
        else{
            binaryTreePathsDFS(root.left, path+"->",ans);
            binaryTreePathsDFS(root.right, path+"->",ans);
        }
    }
}