<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>

class Solution
{
private:
    map<string, vector<TreeNode *>> mp;
    string postOrder(TreeNode *root)
    {
        if (!root)
        {
            return "";
        }
        string leftAns = postOrder(root->left);
        string rightAns = postOrder(root->right);
        string currentString = "L" + leftAns + to_string(root->val) + "R" + rightAns;
        mp[currentString].push_back(root);
        return currentString;
    }

public:
    vector<TreeNode *> findDuplicateSubtrees(TreeNode *root)
    {
        postOrder(root);
        vector<TreeNode *> ans;
        for (auto [key, value] : mp)
        {
            if (value.size() > 1)
            {
                ans.push_back(value[0]);
            }
        }
        return ans;
    }
};
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