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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func binaryTreePaths(root *TreeNode) (ans []string) {
	t := []string{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		t = append(t, strconv.Itoa(root.Val))
		if root.Left == nil && root.Right == nil {
			ans = append(ans, strings.Join(t, "->"))
		} else {
			dfs(root.Left)
			dfs(root.Right)
		}
		t = t[:len(t)-1]
	}
	dfs(root)
	return
}