<p>Given the <code>root</code> of a binary tree, return <em>all root-to-leaf paths in <strong>any order</strong></em>.</p>

<p>A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/paths-tree.jpg" style="width: 207px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,5]
<strong>Output:</strong> [&quot;1-&gt;2-&gt;5&quot;,&quot;1-&gt;3&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [&quot;1&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>


class Solution {
public:
    void dfs(TreeNode* root, string temp, vector<string>& ans){
        // bc
        if(!root) return;
        if(!root->left && !root->right){
            temp+= to_string(root->val);
            ans.push_back(temp);

        }

        //task
        temp+= to_string(root->val) + "->";
        // rr
        dfs(root->left, temp, ans);
        dfs(root->right, temp, ans);

        
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ans;
        string temp="";
        dfs(root, temp, ans);
        return ans;

        
    }
};
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