leetcode刷题日记——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 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],

[15,7]

问题分析：题目所要结果就是二叉数的广度优先搜索的遍历。但是题目目标和一般的广度优先搜索略有区别，就是需要把遍历的序列按照层次关系保存起来，所以单纯的用一个队列来实现，并不能实现题目的目标，这里采用了一个辅助的队列，然后交替的去保存每一层的元素，然后再交替访问二个队列，将数据以层次关系保存到数组中。实现代码如下：

/** * 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<vector<int>> travel;        queue<TreeNode*> duilie1,duilie2;        if(root==NULL) return travel;        duilie1.push(root);        vector<int> aa;        while((!duilie1.empty())||(!duilie2.empty())){             if(!duilie1.empty()){               while(!duilie1.empty()){                   TreeNode *p=duilie1.front();                   aa.push_back(p->val);                   duilie1.pop();                   if(p->left!=NULL)                       duilie2.push(p->left);                   if(p->right!=NULL)                       duilie2.push(p->right);               }                travel.push_back(aa);                aa.clear();             }             if(!duilie2.empty()){               while(!duilie2.empty()){                   TreeNode *p=duilie2.front();                   aa.push_back(p->val);                   duilie2.pop();                   if(p->left!=NULL)                       duilie1.push(p->left);                   if(p->right!=NULL)                       duilie1.push(p->right);               }               travel.push_back(aa);               aa.clear();             }        }        return travel;    }};