---
tags: leetcode
---

# [652. Find Duplicate Subtrees](https://leetcode.com/problems/find-duplicate-subtrees/)
---
# My Solution
## The Key Idea for Solving This Coding Question
DFS (recursion, postorder traversal) + hash map
## C++ Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    vector<TreeNode *> findDuplicateSubtrees(TreeNode *root) {
        vector<TreeNode *> answer;
        unordered_map<string, int> subtreeCnt;

        dfs(root, subtreeCnt, answer);
        return answer;
   }

private:
    string dfs(TreeNode *root, unordered_map<string, int> &subtreeCnt, 
               vector<TreeNode *> &answer) {
        if (root == nullptr) {
            return "x,";
        }

        string subtree = dfs(root->left, subtreeCnt, answer) + 
                         dfs(root->right, subtreeCnt, answer) + 
                         to_string(root->val) + ",";
        auto iter = subtreeCnt.find(subtree);
        if (iter == subtreeCnt.end()) {
            subtreeCnt[subtree] = 1;
        } else {
            if (iter->second == 1) {
                answer.push_back(root);
            }
            ++iter->second;
        }
        return subtree;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the binary tree referred by `root`.
## Space Complexity
$O(H)$
$H$ is the height of the binary tree referred by `root`.
# Miscellaneous
<!--
# Test Cases

[1,2,3,4,null,2,4,null,null,4]

[2,1,1]

[2,2,2,3,null,3,null]

[1,2,2,3,4,3,4,5,null,6,null,5,null,null,6]

[2,1,3]

* wrong answer:

[10,2,22,1,12,1,1]

* wrong answer:

[0,0,0,0,null,null,0,0,0,0,0]

-->