Description:
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/
9 20
/
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
分析： 题目要给出二叉树的zigzag Z型遍历，其本质应该是二叉树的宽搜，然后根据层数来确定每一层是否翻转。
这主要点就在于宽搜时需要知道当前在第几层，所以需要两个queue来做。 这个是一个常用技巧
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 13 vector<vector<int> >result; 14 if(root==NULL) return result; 15 16 queue<TreeNode* > q,qadd; 17 q.push(root); 18 vector<int> level; 19 int nowlevel = 1; 20 while(!q.empty()) 21 { 22 TreeNode *one = q.front(); 23 level.push_back(one->val); 24 if(one->left!=NULL) qadd.push(one->left); 25 if(one->right!=NULL) qadd.push(one->right); 26 27 q.pop(); 28 if(q.empty()) 29 { 30 if(nowlevel%2==0) reverse(level.begin(),level.end()); 31 result.push_back(level); 32 level.clear(); 33 nowlevel++; 34 swap(q,qadd); 35 } 36 } 37 return result; 38 } 39 };