leetcode 103. Binary Tree Zigzag Level Order Traversal 二叉树的锯齿形层次遍历

与102相比就增加了flag，用以确定要不要进行reverse操作

reverse：STL公共函数，对于一个有序容器的元素reverse ( s.begin(),s.end() )可以使得容器s的元素顺序反转；

C++代码：

 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  */
10 class Solution {
11 public:
12     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
13         if(root==NULL) return {};
14         queue<TreeNode*> q;
15         TreeNode* front;
16         q.push(root);
17         vector<vector<int>> res;
18         int flag=-1;
19         while(!q.empty()){
20             flag=-flag;
21             vector<int> onelevel;
22             for(int i=q.size();i>0;i--){
23                 front=q.front();
24                 q.pop();
25                 if(front->left)
26                     q.push(front->left);
27                 if(front->right)
28                     q.push(front->right);
29                 onelevel.push_back(front->val);
30             }
31             if(flag<0) reverse(onelevel.begin(),onelevel.end());
32             res.push_back(onelevel);
33         }
34         return res;

 相似：

102 103 107