# Using a count to mark the level we can slove this with Level Order Traversal easily

• 1.this problem is similarity with Binary Tree Level Order Traversal ,we can add a count to mark the level
2.when level % 2 == 0,we reverse the number in the vector,then push them into the vector result.other else we push them into the vector straightly.

``````class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
queue<TreeNode *> q;
if (!root)
{
return result;
}
q.push(root);
q.push(NULL);
vector<int> cur;
int level = 1;
while (!q.empty())
{
TreeNode* t = q.front();
q.pop();
if (t == NULL)
{
if (level % 2 == 0)
{
reverse(cur.begin(), cur.end());
}
result.push_back(cur);
cur.resize(0);
if (q.size() > 0)
{
q.push(NULL);
}
level++;
}
else
{
cur.push_back(t->val);
if (t->left)
{
q.push(t->left);
}
if (t->right)
{
q.push(t->right);
}
}
}
return result;
}
};
``````

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