103.Binary Tree Zigzag Level Order Traversal

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]
]

思路：层序遍历，利用队列，一次性输出一层的所有节点，如果是偶数层，就将这一层输出的数组反转。

1. /**
    * 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 {
   private:
       vector<vector<int>> result;
   public:
       vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
           if(!root)
               return result;
           queue<TreeNode *> q;
           q.push(root);
           int dep=1;
           while(!q.empty()){
               int size=q.size();
               vector<int> level;
               for(int i=0;i<size;i++){
                   TreeNode * node=q.front();
                   q.pop();
                   level.push_back(node->val);
                   if(node->left)
                       q.push(node->left);
                   if(node->right) 
                       q.push(node->right);
               }
               if(dep%2==0)
                   reverse(level.begin(),level.end());
               dep++;
               result.push_back(level);
           }
           return result;
       }
   };