257. Binary Tree Paths

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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"]

二叉树的遍历，下面这个代码是我参考网上其他大神的代码写的。

这个代码真的写的很赞！短小，精练！比我自己纠结很久写的臭长的代码好很多！/** * 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> result;    void getDfsPaths( TreeNode* node, string strpath) {                if(!node->left && !node->right){            result.push_back(strpath);            return ;        }        if(node->left)            getDfsPaths(node->left, strpath+"->"+to_string(node->left->val));        if(node->right)            getDfsPaths( node->right, strpath+"->"+to_string(node->right->val));    }    vector<string> binaryTreePaths(TreeNode* root) {                if(!root)             return result;                getDfsPaths( root, to_string(root->val));        return result;    }};

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