# Two C++ Solutions, both not recursive

• solution:

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

solution1:

``````    vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root) return res;
queue<pair<TreeNode* , int>> q;
q.push(make_pair(root, 0));
while(!q.empty()) {
TreeNode* node = q.front().first;
int level = q.front().second;
q.pop();
if(level == res.size()) res.push_back( vector<int>() );
res[level].push_back(node->val);
if(node->left) q.push(make_pair(node->left, level + 1));
if(node->right) q.push(make_pair(node->right, level + 1));

}
return res;
}
``````

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