# C++. 8ms Iterative (queue+stack)

• ``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> nodes;
if(!root)
return nodes;

stack<vector<int>> s;
queue<TreeNode*> q;
q.push(root);

while(1){
int nodecount = q.size();
vector<int> levelnodes;
if(nodecount == 0)
break;

while(nodecount > 0){
TreeNode *p = q.front();
q.pop();
levelnodes.push_back(p->val);
nodecount--;
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
}
// one level complete
s.push(levelnodes);
}
// now push elements to the nodes vector
while(s.size()){
nodes.push_back(s.top());
s.pop();
}

return nodes;

}
};`
``````

May be this solution has been discussed before.
We can use stack to reverse the elements.

• this algorithm is not so good;
I don't want extra space to store the vector<int>

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