Binary Tree Paths
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0257.Binary%20Tree%20Paths/images/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>
Tree; Depth-First Search; String; Backtracking; Binary Tree
/**
 * 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> ans;
        vector<string> t;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            t.push_back(to_string(root->val));
            if (!root->left && !root->right) {
                ans.push_back(join(t));
            } else {
                dfs(root->left);
                dfs(root->right);
            }
            t.pop_back();
        };
        dfs(root);
        return ans;
    }

    string join(vector<string>& t, string sep = "->") {
        string ans;
        for (int i = 0; i < t.size(); ++i) {
            if (i > 0) {
                ans += sep;
            }
            ans += t[i];
        }
        return ans;
    }
};