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]
]
要注意读取的顺序，是折线。每次都应该按照不同的顺序。

/**
 * 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> > re;
    vector<vector <TreeNode *> > t;
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        if( root != NULL)
        {
            vector<int> ve;
            ve.push_back(root->val);
            vector <TreeNode *> tr;
            tr.push_back(root->right);
            tr.push_back(root->left);
            re.push_back(ve);
            t.push_back(tr);
            get(0);
        }
        
        return re;
    }
    void get(int n)
    {
        vector<int> ve;
        vector <TreeNode *> tr;
        if(n%2 == 0)
        {
            vector<TreeNode *> tt;
            for(int i = 0 ; i < t[n].size();i++)
            {
                if(t[n][i] != NULL)
                {
                    ve.push_back(t[n][i]->val);
                    tt.push_back(t[n][i]->right);
                    tt.push_back(t[n][i]->left);
                }
            }
            for(int i = tt.size()-1;i>=0;i--)
            tr.push_back(tt[i]);
            
        }
        else{
            vector<TreeNode *> tt;
            for(int i = 0 ; i < t[n].size();i++)
            {

                if(t[n][i] != NULL)
                {
                    ve.push_back(t[n][i]->val);
                    tt.push_back(t[n][i]->left);
                    tt.push_back(t[n][i]->right);
                }
            }
            for(int i = tt.size()-1;i>=0;i--)
                tr.push_back(tt[i]);
        }
        
        if(ve.size()==0)return;
        
        re.push_back(ve);
        t.push_back(tr);
        get(n+1);
        return ;
    }
};