102. Binary Tree Level Order Traversal

O(N) Time and O(N) Space

• Create an empty queue q.
• Push the root node of tree to q.
• Loop while the queue is not empty:
  • get all the elements of q.
  • push their left and right nodes in the queue.
  • push_back these elements in the vector.
  • push_back this vector in main 2d vector.
• return 2d vector.

Code

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> levelorder;
        if(root==NULL)return levelorder;

        queue<TreeNode *> q;
        q.push(root);

        while(!q.empty()){
            int sz=q.size();
            vector<int> level;
            for(int i=0;i<sz;i++){
                TreeNode *node= q.front(); q.pop();
                if(node->left!=NULL)q.push(node->left);
                if(node->right!=NULL)q.push(node->right);
                level.push_back(node->val);
            }
            levelorder.push_back(level);
        }
        return levelorder;
    }
};