---
# System prepended metadata

title: 257. Binary Tree Paths
tags: [dfs, tree, Leetcode, python, c++, easy]

---

###### tags: `Leetcode` `easy` `tree` `dfs` `python` `c++`

# 257. Binary Tree Paths

## [題目連結:] https://leetcode.com/problems/binary-tree-paths/description/

## 題目:
Given the ```root``` of a binary tree, return all root-to-leaf paths in **any order**.

A **leaf** is a node with no children.

![](https://i.imgur.com/iBYwCLd.png)
![](https://i.imgur.com/OSKwv9a.png)


## 解題想法:
* 此題為輸出二元樹的所有路徑，並以"->"串接
* DFS遞歸即可:
    * Sol1: 直接用原函式遞迴
        * 初始需判斷是否為空or沒左右子
        * 遞迴左右子創建path
        * 將當前root分別與遞迴左右子的結果串接
        * merge兩段即為解
    * Sol2: 額外寫DFS遞迴
        * 原函式給input:
            * path=""
            * res=[] (或可以用self.res)
        * dfs
            * 初始判斷root是否為空:
                * 空則直接return nothinh
            * path+=str(root.val)
            * 再判斷是否左右子皆空:
                * 若皆空則將該path加入res並return
            * 遞迴左子:
                * dfs(root.left, path+"->")
            * 遞迴右子:
                * dfs(root.right, path+"->")

## Python: Sol1
``` python=
class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        if not root.left and not root.right:
            return [str(root.val)]
        leftPath=self.binaryTreePaths(root.left)
        rightPath=self.binaryTreePaths(root.right)
        #字串包在一起 用+  ex: '1'+'->'+'2' = '1->2'
        leftMerge=[str(root.val)+"->"+ val for val in leftPath]
        rightMerge=[str(root.val)+"->"+val for val in rightPath]
        return leftMerge+rightMerge
```

## Python: Sol2
``` python=
class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        self.res=[]
        path=""
        self.dfs(root, path)
        return self.res
    def dfs(self, root, path):
        if not root:
            return 
        path+=str(root.val)
        if not root.left and not root.right:
            self.res.append(path)
            return
        if root.left:
            self.dfs(root.left, path+"->")
        if root.right:
            self.dfs(root.right, path+"->")
        return 
```

## C++: Sol1:
``` cpp=
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        if (!root) return {};
        if (!root->left && !root->right) return {to_string(root->val)};
        vector<string> LeftPath, RightPath;
        LeftPath=binaryTreePaths(root->left);
        RightPath=binaryTreePaths(root->right);
        //merge LeftPath and RightPath
        for (string item: RightPath){
            LeftPath.push_back(item);
        }
        //add "root->val"+"->" in front of all string in LeftPath
        for (string& item: LeftPath){
            item= to_string(root->val)+"->"+item;
        }
        return LeftPath;
    }
};
```

## C++: Sol2
``` cpp=
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        string path="";
        dfs(root, res, path);
        return res;
    }
    void dfs(TreeNode* root, vector<string>& res, string path){
        if (root==nullptr) return;
        path+=to_string(root->val);
        if (!root->left && !root->right){
            res.push_back(path);
            return;
        }
        if (root->left) dfs(root->left, res, path+"->");
        if (root->right) dfs(root->right, res, path+"->");
        return ;
    }
};
```