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Utilize a Depth-First Search (DFS) strategy and serialize each subtree. Treat the serialization as a key in a hashmap to track occurrences of identical subtrees. Duplicate entries are detected when a serialized key is encountered more than once.
Time Complexity: O(N), where N is the number of nodes since we visit each node once.
Space Complexity: O(N), accounting for the hashmap to store serialized trees and recursion stack space.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4#include <unordered_map>
5#include <vector>
6
7struct TreeNode {
8 int val;
9 struct TreeNode *left;
10 struct TreeNode *right;
11};
12
13char* serialize(struct TreeNode* node, std::unordered_map<std::string, std::vector<struct TreeNode*>>& map) {
14 if (!node) return "#";
15 char* left = serialize(node->left, map);
16 char* right = serialize(node->right, map);
17 std::string serial = std::to_string(node->val) + "," + std::string(left) + "," + std::string(right);
18 map[serial].push_back(node);
19 return strdup(serial.c_str());
20}
21
22std::vector<struct TreeNode*> findDuplicateSubtrees(struct TreeNode* root) {
23 std::unordered_map<std::string, std::vector<struct TreeNode*>> map;
24 serialize(root, map);
25 std::vector<struct TreeNode*> result;
26 for (auto& pair : map) {
27 if (pair.second.size() > 1) {
28 result.push_back(pair.second.front());
29 }
30 }
31 return result;
32}This C solution serializes the tree into strings using recursive DFS. A hashmap is employed to track these serialized strings and identify duplicates. Nodes with duplicate serials are returned.
Rather than using direct serialization strings, assign each unique subtree encountered a unique ID using a hash map. If a subtree's ID is seen again, it's a duplicate.
Time Complexity: O(N), as we traverse each node.
Space Complexity: O(N), necessary for dictionaries and recursive processing space.
1#include <unordered_map>
2#include <vector>
3
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
unordered_map<string, int> ids;
unordered_map<string, int> count;
vector<TreeNode*> result;
int idCounter = 1;
public:
vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
serialize(root);
return result;
}
private:
int serialize(TreeNode* node) {
if (!node) return 0;
string serial = to_string(node->val) + ',' + to_string(serialize(node->left)) + ',' + to_string(serialize(node->right));
if (!ids.count(serial)) {
ids[serial] = idCounter++;
}
int id = ids[serial];
if (++count[serial] == 2) {
result.push_back(node);
}
return id;
}
};This C++ implementation stands by using unique identifiers to serialize subtree structures uniquely. Duplicate subtrees' roots are identified through the repeated presence of the generated IDs.