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Given the root
of a binary tree, return all duplicate subtrees.
For each kind of duplicate subtrees, you only need to return the root node of any one of them.
Two trees are duplicate if they have the same structure with the same node values.
Example 1:
Input: root = [1,2,3,4,null,2,4,null,null,4]
Output: [[2,4],[4]]
Example 2:
Input: root = [2,1,1]
Output: [[1]]
Example 3:
Input: root = [2,2,2,3,null,3,null]
Output: [[2,3],[3]]
Constraints:
[1, 10^4]
-200 <= Node.val <= 200
class Solution {
public:
vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
unordered_map<string, vector<TreeNode*>> map;
vector<TreeNode*> dups;
serialize(root, map);
for (auto it = map.begin(); it != map.end(); it++)
if (it->second.size() > 1) dups.push_back(it->second[0]);
return dups;
}
private:
string serialize(TreeNode* node, unordered_map<string, vector<TreeNode*>>& map) {
if (!node) return "";
string s = "(" + serialize(node->left, map) + to_string(node->val) + serialize(node->right, map) + ")";
map[s].push_back(node);
return s;
}
};
class Solution {
public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
Map<String, List<TreeNode>> map = new HashMap<String, List<TreeNode>>();
List<TreeNode> dups = new ArrayList<TreeNode>();
serialize(root, map);
for (List<TreeNode> group : map.values())
if (group.size() > 1) dups.add(group.get(0));
return dups;
}
private String serialize(TreeNode node, Map<String, List<TreeNode>> map) {
if (node == null) return "";
String s = "(" + serialize(node.left, map) + node.val + serialize(node.right, map) + ")";
if (!map.containsKey(s)) map.put(s, new ArrayList<TreeNode>());
map.get(s).add(node);
return s;
}
}
def findDuplicateSubtrees(self, root):
def trv(root):
if not root: return "null"
struct = "%s,%s,%s" % (str(root.val), trv(root.left), trv(root.right))
nodes[struct].append(root)
return struct
nodes = collections.defaultdict(list)
trv(root)
return [nodes[struct][0] for struct in nodes if len(nodes[struct]) > 1]
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