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

 

 /**
  * 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> > zigzagLevelOrder(TreeNode *root) {
         vector<vector<int>> res;
         if(root==NULL) return res;
         queue<TreeNode*> q;
         q.push(root);
         bool reverse=false;
         while(!q.empty()){
             vector<int> v;
             int size=q.size();
             for(int i=;i<size;++i){
                 TreeNode *cur=q.front();
                 q.pop();
                 v.push_back(cur->val);
                 if(cur->left!=NULL) q.push(cur->left);
                 if(cur->right!=NULL) q.push(cur->right);
             }
             if(reverse){

                 vector<int> tmp;
                 for(int i=v.size()-;i>=;--i){
                     tmp.push_back(v[i]);
                 }
                 res.push_back(tmp);
             }else{
                 res.push_back(v);
             }
             reverse=!reverse;
         }
         return res;
     }
 };