<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]]:
        #binary tree
        #same structure, same node values
        #1.graph: key=values, value=list of node
        #2.check the graph, compare each node?
        #node inorder traversal: print the subreee with inorder traversal, the string use the string as key.
        #node is value
        graph=dict()
        result=list()
        self.helper(root,graph,result)
        return result


    def helper(self,node,graph,result):
        if node==None:
            return ""

        left=self.helper(node.left,graph,result)
        right=self.helper(node.right,graph,result)
        #left and right should add extra ()
        cur="("+left+")"+"("+right+")"+"("+str(node.val)+")"

        if cur in graph:
            graph[cur]=graph[cur]+1
        else:
            graph[cur]=1

        if graph[cur]==2:
            result.append(node)
            print(result)

        return cur
        
29348