"Leetcode" Binary Tree level Order traversal

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    /   7

Return its level order traversal as:

[  3],  [9,20],  [15,7]]

Idea: Use a queue for storage, using two variables at the same time to record the number of nodes in that layer, and the number of nodes in the next layer.

/** * Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * Tre Enode (int x): Val (x), left (null), right (NULL) {} *}; */class Solution {public:vector<vector<int> > Levelorder (TreeNode *root) {vector<vector<int&                gt;> result;                if (root = NULL) return result;        Queue<treenode *> LQ;        Lq.push (root);        int CURLC = 1;                int NEXTLC = 0;            while (!lq.empty ()) {vector<int> level;                while (CURLC > 0) {TreeNode * temp = Lq.front ();                Lq.pop ();                curlc--;                                Level.push_back (Temp->val);                    if (temp->left) {Lq.push (temp->left);                nextlc++;             } if (temp->right) {Lq.push (temp->right);       nextlc++;            }} CURLC = NEXTLC;            NEXTLC = 0;        Result.push_back (level);    } return result; }};

"Leetcode" Binary Tree level Order traversal