(Leetcode) Binary Tree level Order traversal

Binary Tree level Order Traversalmy SubmissionsQuestionSolutionTotal Accepted £ º 67360 Total submissions: 228717 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    /   7

Return its level order traversal as:

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

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

1 /**2 * Definition for a binary tree node.3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8  * };9  */Ten classSolution { One  Public: Avector<vector<int>> Levelorder (treenode*root) { -Queue<treenode*>que; -vector<vector<int>>ret; the         if(!root)returnret; -         intCur,next; - Que.push (root); -Cur =1; next =0; +vector<int>tmp; -          while(!que.empty ()) +         { Atreenode*Tmpnode; atTmpnode =Que.front (); -Tmp.push_back (tmpnode->val); - Que.pop (); -cur--; -             if(Tmpnode->left) {Que.push (tmpnode->left); next++; } -             if(Tmpnode->right) {Que.push (tmpnode->right); next++; } in             if(cur = =0{cur = next; next =0; Ret.push_back (TMP); Tmp.clear ();} -              to         } +          -         returnret; the     } *};

(Leetcode) Binary Tree level Order traversal