Leetcode刷题笔记题解（C++）：257. 二叉树的所有路径
思路：深度优先搜索
/*** 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) {//保存结果vector<string> res;//进行DFSdfs(root,"",res);return res;}void dfs(TreeNode* root,string path,vector<string>& res){//当前节点不为空if(root!=nullptr){//加上节点的值path += to_string(root->val);//如果当前节点为叶子节点则到头了加入结果并返回if(root->left == nullptr && root->right == nullptr){res.push_back(path);return;}//当前节点不为空的话则需要向下搜索，左子树和右子树继续深度搜索path += "->";dfs(root->left,path,res);dfs(root->right,path,res);}}
};