[LeetCode]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}".

思考：偶数层直接反转，竟然过了。

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

private:

	vector<vector<int> > ret;

public:

    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

		ret.clear();

		if(root==NULL) return ret;

		vector<int> ans;

		queue<TreeNode *> q;	

		q.push(root);

		int num=1;

		while(!q.empty())

		{

			ans.clear();

			int count=0;

			while(num--)

			{			

				TreeNode *node=q.front();

				q.pop();

				ans.push_back(node->val);

				if(node->left)

				{

					count++;

					q.push(node->left);

				}

				if(node->right)

				{

					count++;

					q.push(node->right);

				}

			}

			num=count;

			ret.push_back(ans);

		}

		for(int i=0;i<ret.size();i++)

		{

			if(i%2==0) continue;

			for(int j=0;j<ret[i].size()/2;j++)

				swap(ret[i][j],ret[i][ret[i].size()-j-1]);

		}

		return ret;

	}

};