## [257\. Binary Tree Paths](https://leetcode.com/problems/binary-tree-paths/)

Given the `root` of a binary tree, return _all root-to-leaf paths in **any order**_.

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

**Example 1:**


**Input:** root = \[1,2,3,null,5\]
**Output:** \["1->2->5","1->3"\]

**Example 2:**

**Input:** root = \[1\]
**Output:** \["1"\]

**Constraints:**

- The number of nodes in the tree is in the range `[1, 100]`.
- `-100 <= Node.val <= 100`

recursion

/**
 * 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) {
        vector<string> res;
        dfs(root, "", res);
        return res;    
    }

    void dfs(TreeNode* root, string out, vector<string>& res) {
        if(!root->left && !root->right) {
            res.push_back(out + to_string(root->val));
        }
        if(root->left) {
            dfs(root->left, out + to_string(root->val) + "->", res);
        }
        if(root->right) {
            dfs(root->right, out + to_string(root->val) + "->", res);
        }        
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::

iteration

/**
 * 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) {
        vector<string> res;
        // base case
        if(!root) return res;
        
        // 如果沒有左右子節點，則返回 root
        if(!root->left && !root->right) {
            return {to_string(root->val)};
        }
        
        // 找左側
        for(auto s : binaryTreePaths(root->left)) {
            res.push_back(to_string(root->val) + "->" + s);
        }
        
        // 找右側
        for(auto s : binaryTreePaths(root->right)) {
            res.push_back(to_string(root->val) + "->" + s);
        }

        return res;        
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::