<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 buildString(vector<int> &backtrack, vector<string> &ans) {
        string ans1 = "";
        for (int i = 0; i < backtrack.size(); i++) {
            ans1 += to_string(backtrack[i]);
            if (i != backtrack.size() - 1) {
                ans1 += "->";
            }
        }
        ans.push_back(ans1);
    }

    void backtrack(TreeNode* root, vector<int> &currentPath, vector<string> &ans) {
        if (root == nullptr) {
            return;
        }
        currentPath.push_back(root->val);
        if (root->left == nullptr && root->right == nullptr) {
            buildString(currentPath, ans);
        } else {
            backtrack(root->left, currentPath, ans);
            backtrack(root->right, currentPath, ans);
        }
        currentPath.pop_back();
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ans;
        vector<int> currentPath;
        backtrack(root, currentPath, ans);
        return ans;
    }
};

16400