给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

输入:

   1/   \
2     3\5

输出: [“1->2->5”, “1->3”]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

代码如下:

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<string> binaryTreePaths(TreeNode* root) {auto paths = vector<string>();struct_paths(root,"",paths);return paths;}void struct_paths(TreeNode *root,string path,vector<string>&paths){if (root!=nullptr){path+=to_string(root->val);if (root->left==nullptr && root->right==nullptr){paths.push_back(path);}else {path+="->";struct_paths(root->left,path,paths);struct_paths(root->right,path,paths);}}}
};