【leetcode】Binary Tree Level Order Traversal

Question ：

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]]

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

Anwser 1 ：         

/** * Definition for binary tree * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int>> ret;                if(root == NULL) return ret;                vector<int> vec;        queue<TreeNode *> Q;                Q.push(root);        int count = 1;                while(!Q.empty()){                        vec.clear();            int nextCount = 0;      // cal next row count            for(int i = 0; i < count; i++){     // one row count                TreeNode *tmp = Q.front();                Q.pop();                                vec.push_back(tmp->val);        // save one row val                if(tmp->left){                    Q.push(tmp->left);                    nextCount++;                }                if(tmp->right){                    Q.push(tmp->right);                    nextCount++;                }            }            count = nextCount;            ret.push_back(vec);        }        return ret;    }};

Anwser 2 ：           

/** * Definition for binary tree * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int>> ret;        if(root == NULL) return ret;                vector<int> vec;        queue<TreeNode *> Q;        queue<TreeNode *> Q2;   // extra space                Q.push(root);        while(!Q.empty()){            TreeNode *tmp = Q.front();            Q.pop();                        if(tmp != NULL){                vec.push_back(tmp->val);                if(tmp->left) Q2.push(tmp->left);                if(tmp->right) Q2.push(tmp->right);            }                        if(Q.empty()){      // one row end                ret.push_back(vec);                vec.clear();                swap(Q, Q2);            }        }    }};