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]
]

 

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}".

思路：这道题关键有两点：一是辅助空间的选择——队列，二是标志符号——第一层从左到右，第二层从右到左，第三层从左到右......，flag为0时表示从左到右，为1时从右到左。用vector<int> temp来存储每一层的结点值，如果flag==1，则逆转temp，然后存入result里，如果flag==0，直接存入result；

/**
 * 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> > result;
        vector<int> temp;
        result.clear();
        if(root==NULL)
            return result;
        queue<TreeNode *> pTemp;
        int flag=0;
        pTemp.push(root);
        pTemp.push(NULL);
        while(!pTemp.empty())
        {
            TreeNode *curNode=pTemp.front();
            pTemp.pop();
            if(curNode!=NULL)
            {
                temp.push_back(curNode->val);
                if(curNode->left)
                    pTemp.push(curNode->left);
                if(curNode->right)
                    pTemp.push(curNode->right);
            }
            else
            {
                if(!temp.empty())
                {
                    pTemp.push(NULL);
                    if(flag==1)
                    {
                        reverse(temp.begin(),temp.end());
                    }
                    result.push_back(temp);
                    flag=1-flag;
                    temp.clear();
                }
            }
        }
        return result;
    }
};