<p>Given the roots of two binary trees <code>root</code> and <code>subRoot</code>, return <code>true</code> if there is a subtree of <code>root</code> with the same structure and node values of<code> subRoot</code> and <code>false</code> otherwise.</p>

<p>A subtree of a binary tree <code>tree</code> is a tree that consists of a node in <code>tree</code> and all of this node&#39;s descendants. The tree <code>tree</code> could also be considered as a subtree of itself.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg" style="width: 532px; height: 400px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,2], subRoot = [4,1,2]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg" style="width: 502px; height: 458px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the <code>root</code> tree is in the range <code>[1, 2000]</code>.</li>
	<li>The number of nodes in the <code>subRoot</code> tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>-10<sup>4</sup> &lt;= root.val &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= subRoot.val &lt;= 10<sup>4</sup></code></li>
</ul>

# 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
from collections import deque
class Solution:
    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:

        def sameSubtree(root1, root2):
            if not root1 and not root2:
                return True
            elif not root1 or not root2 or root1.val != root2.val:
                return False
            else:
                return sameSubtree(root1.left, root2.left) and sameSubtree(root1.right, root2.right)

        # BFS search the root tree, if the value is equal to the root of subtre -> call sameSubtree
        if not root or not subRoot:
            return False

        queue = deque([root])
        while queue:
            for _ in range(len(queue)):
                node = queue.popleft()
                if node.val == subRoot.val and sameSubtree(node, subRoot):
                    return True
                
                # add children to queue
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

        return False

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