<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]]:
        seen = set()
        dup = set()
        def recur(u):
            if u is None:
                return ""
            s = str(u.val) + "(" + recur(u.left) + ")(" + recur(u.right) + ")"
            if s in seen:
                dup.add(s)
            else:
                seen.add(s)
            return s
        recur(root)
        # print(dup)
        
        def process(s, i, j):
            if j-i<=1:
                return None
            v = None
            stk = []
            l, r = -1, -1
            for k in range(i, j):
                if s[k] == "(":
                    if v is None:
                        v = int(s[i:k])
                    stk.append(k)
                elif s[k] == ")":
                    idx_open = stk.pop()
                    if len(stk) == 0:
                        if l == -1:
                            l = process(s, idx_open+1, k)
                        else:
                            r = process(s, idx_open+1, k)
                            break
            return TreeNode(v, l, r)
        
        return [process(x, 0, len(x)) for x in dup]

#         seen = set()
#         dup = set()
#         def recur(u):
#             if u is None:
#                 return ""
#             s = str(u.val) + "(" + recur(u.left) + ")(" + recur(u.right) + ")"
#             if s in seen:
#                 dup.add(s)
#             else:
#                 seen.add(s)
#             return s
#         recur(root)
#         print(dup)
        
#         def deflate(s):
            
            
        # result = []
        # for tree_str in dup:
        #     result.append(deflate(tree_str))
        # return result
        
            
                
        
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