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652. Find Duplicate Subtrees 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
    vector<TreeNode*> ans;
    unordered_map<string, int> count;
    encode(root, count, ans);
    return ans;
  }

 private:
  string encode(TreeNode* root, unordered_map<string, int>& count,
                vector<TreeNode*>& ans) {
    if (root == nullptr)
      return "";

    const string encoded = to_string(root->val) + "#" +
                           encode(root->left, count, ans) + "#" +
                           encode(root->right, count, ans);
    if (++count[encoded] == 2)
      ans.push_back(root);
    return encoded;
  }
};

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class Solution {
  public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
    List<TreeNode> ans = new ArrayList<>();
    Map<String, Integer> count = new HashMap<>();
    encode(root, count, ans);
    return ans;
  }

  private String encode(TreeNode root, Map<String, Integer> count, List<TreeNode> ans) {
    if (root == null)
      return "";

    final String encoded =
        root.val + "#" + encode(root.left, count, ans) + "#" + encode(root.right, count, ans);
    if (count.merge(encoded, 1, Integer::sum) == 2)
      ans.add(root);
    return encoded;
  }
}

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class Solution:
  def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]:
    ans = []
    count = collections.Counter()

    def encode(root: Optional[TreeNode]) -> str:
      if not root:
        return ''

      encoded = str(root.val) + '#' + \
          encode(root.left) + '#' + \
          encode(root.right)
      count[encoded] += 1
      if count[encoded] == 2:
        ans.append(root)
      return encoded

    encode(root)
    return ans