LeetCode 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal
My Submissions QuestionEditorial Solution
Total Accepted: 98904 Total Submissions: 304078 Difficulty: Easy
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]

]

/** * 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>> levelOrder(TreeNode* root) {        vector<int> row;        vector<vector<int>> v;        queue<TreeNode *> q;        if(root == NULL)            return v;        q.push(root);        TreeNode *temp;        while(!q.empty()) {            int size = q.size();            while(size--) {                temp = q.front();                q.pop();                row.push_back(temp->val);                if(temp->left != NULL) {                    q.push(temp->left);                }                if(temp->right != NULL) {                    q.push(temp->right);                }            }            v.push_back(row);            row.clear();        }        return v;    }};