<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>

from collections import deque

class Solution:
    def binaryTreePaths(self, root):
        if not root:
            return []
        
        paths = []
        queue = deque([(root, str(root.val))])  # queue stores (node, path)
        
        while queue:
            node, path = queue.popleft()
            
            # If it's a leaf node, add the path to the result
            if not node.left and not node.right:
                paths.append(path)
            else:
                # Add left child to the queue if it exists
                if node.left:
                    queue.append((node.left, path + '->' + str(node.left.val)))
                # Add right child to the queue if it exists
                if node.right:
                    queue.append((node.right, path + '->' + str(node.right.val)))
        
        return paths

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