<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 {
public:
    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        unordered_map<string, int> found; 
        vector<TreeNode*> dupes;
        stringRep(root, found, dupes);
        return dupes;
    }

    string stringRep(TreeNode* root, unordered_map<string, int>& found, vector<TreeNode*>& dupes){
        if (!root) return "";
        string lst = stringRep(root->left, found, dupes);
        string rst = stringRep(root->right, found, dupes);
        string res;
        if (rst.size() > 0){
            res = to_string(root->val) + "(" + lst + ")"  + "(" + rst + ")";
        }
        else if (lst.size() > 0){
            res = to_string(root->val) + "(" + lst + ")";
        }
        else{
            res = to_string(root->val);
        }
        auto it = found.find(res);
        if (it != found.end()){
            if (it->second == 1){
                dupes.push_back(root);
            }
            it->second++;
        }
        else{
            found[res] = 1;
        }
        return res;
    }
};

//lst rst
//t   t - both
//t   f - left
//f   t -> both
//f   f - NONE

65615