<p>Given the <code>root</code>&nbsp;of a binary tree, return all <strong>duplicate subtrees</strong>.</p>

<p>For each kind of duplicate subtrees, you only need to return the root node of any <b>one</b> of them.</p>

<p>Two trees are <strong>duplicate</strong> if they have the <strong>same structure</strong> with the <strong>same node values</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/16/e1.jpg" style="width: 450px; height: 354px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,null,2,4,null,null,4]
<strong>Output:</strong> [[2,4],[4]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/16/e2.jpg" style="width: 321px; height: 201px;" />
<pre>
<strong>Input:</strong> root = [2,1,1]
<strong>Output:</strong> [[1]]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/16/e33.jpg" style="width: 450px; height: 303px;" />
<pre>
<strong>Input:</strong> root = [2,2,2,3,null,3,null]
<strong>Output:</strong> [[2,3],[3]]
</pre>

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

<ul>
	<li>The number of the nodes in the tree will be in the range <code>[1, 5000]</code></li>
	<li><code>-200 &lt;= Node.val &lt;= 200</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
class Solution:
    def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]:
        self.subtree_set = collections.defaultdict(int)
        self.ans = []
        if root:
            self.dfs(root)
        return self.ans
    
        
    def dfs(self, root):
        if root.left is None and root.right is None: 
            tree_str = "(#)" + str(root.val) + "(#)"
            self.subtree_set[tree_str] += 1
            if self.subtree_set[tree_str] == 2:
                self.ans.append(root)
            return tree_str
                
        left_ts = "#"
        if root.left:
            left_ts = self.dfs(root.left)
        
        left_ts = "(" + left_ts + ")" + str(root.val)
        right_ts = "#"
        if root.right:
            right_ts = self.dfs(root.right)
            
        tree_str = left_ts + "(" + right_ts + ")"
        self.subtree_set[tree_str] += 1
        if self.subtree_set[tree_str] == 2:
            self.ans.append(root)
        return tree_str
        
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