# Short 8ms C++ solution with queue

• ``````class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if (!root) { return {}; }
vector<int> row;
vector<vector<int> > result;
queue<TreeNode*> q;
q.push(root);
int count = 1;

while (!q.empty()) {
if (q.front()->left) { q.push(q.front()->left); }
if (q.front()->right) { q.push(q.front()->right); }
row.push_back(q.front()->val), q.pop();
if (--count == 0) {
result.emplace_back(row), row.clear();
count = q.size();
}
}
return result;
}
};``````

• ``````vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> ans;
queue<TreeNode *> q;
if (root) q.push(root);
while (!q.empty()) {
int len = q.size();
vector<int> level;
for (int i = 0;i < len;++i) {
TreeNode *node = q.front();
level.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
q.pop();
}
ans.push_back(level);
}
return ans;
}``````

• similar solution using 2 loops

``````vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root)
return res;
queue<TreeNode*> q;
q.push(root);
vector<int> v;
while(!q.empty()){
int n=q.size();
while(n>0){
auto node=q.front();
v.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
q.pop();
--n;
}
res.emplace_back(v);
v.clear();
}
return res;
}``````

• Nice solution

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