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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            t.append(str(root.val))
            if root.left is None and root.right is None:
                ans.append("->".join(t))
            else:
                dfs(root.left)
                dfs(root.right)
            t.pop()

        ans = []
        t = []
        dfs(root)
        return ans
