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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.
1import java.util.*;
2
3public class Solution {
4 public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
5 Map<String, List<TreeNode>> map = new HashMap<>();
6 serialize(root, map);
7 List<TreeNode> result = new ArrayList<>();
8 for (List<TreeNode> nodes : map.values()) {
9 if (nodes.size() > 1) {
10 result.add(nodes.get(0));
11 }
12 }
13 return result;
14 }
15
16 private String serialize(TreeNode node, Map<String, List<TreeNode>> map) {
17 if (node == null) {
18 return "#";
19 }
20 String serial = node.val + "," + serialize(node.left, map) + "," + serialize(node.right, map);
21 map.computeIfAbsent(serial, k -> new ArrayList<>()).add(node);
22 return serial;
23 }
24}This Java solution involves recursively serializing each subtree and storing the serialized string in a hashmap. Duplicates are identified by checking the hashmap for strings that occur more than once.
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.