# Binary Tree Paths Problem & Solution

Given the `root` of a binary tree, return all root-to-leaf paths in any order. A leaf is a node with no children.

## C++ Solution

``````#pragma GCC optimize("Ofast")
#pragma GCC optimization("max-inline-insns-recursive-auto")

static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

/**
* 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> results;

binaryTreePaths(root, "", results);

return results;
}

private:
void binaryTreePaths(TreeNode* root, string result, vector<string>& results) {
if (root == nullptr) {
return;
}

if (root->left == nullptr && root->right == nullptr) {
result += to_string(root->val);
results.push_back(result);
}

string tmp = result + to_string(root->val) + "->";

binaryTreePaths(root->left, tmp, results);
binaryTreePaths(root->right, tmp, results);
}
};
``````