Given 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 same node values.
Example 1:
1
/ \
2 3
/ / \
4 2 4
/
4
The following are two duplicate subtrees:
2
/
4
and
4
Therefore, you need to return above trees' root in the form of a list.
题解：
本来用set<TreeNode*>发现没法去重，然后构造了一个JudgeNode结构体
然后。。
第一次ac结果。。448ms
class Solution {
public:
struct JudgeNode {
TreeNode* t;
string s;
};
vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
vector<TreeNode*> ans;
if (root == NULL) {
return ans;
}
vector<TreeNode*> subtree;
map<string, TreeNode*> mp;
getAllSubtrees(root, subtree);
int n = subtree.size();
vector<JudgeNode*> tmp;
for (int i = 0; i < n; i++) {
JudgeNode* j = new JudgeNode();
j->t = subtree[i];
j->s = getString(subtree[i]);
tmp.push_back(j);
}
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (tmp[i]->s == tmp[j]->s) {
mp.insert(pair<string, TreeNode*>(tmp[i]->s, tmp[i]->t));
}
}
}
map<string, TreeNode*>::iterator it;
for (it = mp.begin(); it != mp.end(); it++) {
ans.push_back(it->second);
}
return ans;
}
static string getString (TreeNode* a) {
string res;
queue<TreeNode*> bfs;
bfs.push(a);
res+=a->val;
while (bfs.empty() == false) {
TreeNode* node = bfs.front();
bfs.pop();
if (node->left != NULL) {
bfs.push(node->left);
res += to_string(node->left->val);
}
else {
res += 'x';
}
if (node->right != NULL) {
bfs.push(node->right);
res += to_string(node->right->val);
}
else {
res += 'x';
}
}
return res;
}
static void getAllSubtrees(TreeNode* root, vector<TreeNode*> &subtree) {
queue<TreeNode*> bfs;
bfs.push(root);
while (bfs.empty() == false) {
TreeNode* node = bfs.front();
subtree.push_back(node);
bfs.pop();
if (node->left != NULL) {
bfs.push(node->left);
}
if (node->right != NULL) {
bfs.push(node->right);
}
}
}
};还是低效率的解法。。。我的思路就是这样了。。能优化的都尽可能优化了还是168ms。。。
class Solution {
public:
vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
vector<TreeNode*> subtree;
if (root == NULL) {
return subtree;
}
map<string, vector<TreeNode*>> mp;
getAllSubtrees(root, subtree);
int n = subtree.size();
for (int i = 0; i < n; i++) {
string s = getString(subtree[i]);
if (mp.find(s) != mp.end()) {
mp[s].push_back(subtree[i]);
}
else {
mp.insert(pair<string, vector<TreeNode*>>(s, vector<TreeNode*>(1, subtree[i])));
}
}
subtree.clear();
map<string, vector<TreeNode*>>::iterator it;
for (it = mp.begin(); it != mp.end(); it++) {
if (it->second.size() > 1) {
subtree.push_back(it->second[0]);
}
}
return subtree;
}
static string getString (TreeNode* a) {
string res;
queue<TreeNode*> bfs;
bfs.push(a);
res += a->val;
while (bfs.empty() == false) {
TreeNode* node = bfs.front();
bfs.pop();
if (node->left != NULL) {
bfs.push(node->left);
res += to_string(node->left->val);
}
else {
res += 'x';
}
if (node->right != NULL) {
bfs.push(node->right);
res += to_string(node->right->val);
}
else {
res += 'x';
}
}
return res;
}
static void getAllSubtrees(TreeNode* root, vector<TreeNode*> &subtree) {
queue<TreeNode*> bfs;
bfs.push(root);
while (bfs.empty() == false) {
TreeNode* node = bfs.front();
subtree.push_back(node);
bfs.pop();
if (node->left != NULL) {
bfs.push(node->left);
}
if (node->right != NULL) {
bfs.push(node->right);
}
}
}
};