binary-tree-zigzag-level-order-traversal——二叉树分层输出

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]
]

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

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

PS：二叉树分层输出，BFS并且记录翻转情况

 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>> res;
14         if(root==NULL) return res;
15         queue<TreeNode*> q;
16         q.push(root);
17         bool reverse=false;
18         while(!q.empty()){
19             vector<int> v;
20             int size=q.size();
21             for(int i=0;i<size;++i){
22                 TreeNode *cur=q.front();
23                 q.pop();
24                 v.push_back(cur->val);
25                 if(cur->left!=NULL) q.push(cur->left);
26                 if(cur->right!=NULL) q.push(cur->right);
27             }
28             if(reverse){
29
30                 vector<int> tmp;
31                 for(int i=v.size()-1;i>=0;--i){
32                     tmp.push_back(v[i]);
33                 }
34                 res.push_back(tmp);
35             }else{
36                 res.push_back(v);
37             }
38             reverse=!reverse;
39         }
40         return res;
41     }
42 };