leecode 解题总结：103. Binary Tree Zigzag Level Order Traversal

#include <iostream>
#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;
/*
问题:
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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]

分析:此题是在之前层序遍历的基础上修改为z型方向遍历，即从左到右，从右到左，然后从左到右这种
难点在于如何修改遍历的顺序，队列先进先出。
一种简单的方法是：
举例:
    1
   2 3
  4 5 6 7
 8 9  10  11
 14 13 12
 那么结果是:
 [1],
 [3,2]
 [4，5,6,7]
 也就是说：第一行是顺序，第2行是逆序，第3行是顺序，第4行数逆序，。。。
 
 关键:
 1 第奇数行是顺序，第偶数行是逆序，仍然用队列+层序遍历方式，
 只不过生成结果后，如果当前是第偶数行，就需要将该行结果逆置
*/

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>> results;
        if(!root)
		{
			return results;
		}
		queue<TreeNode*> nodes;
		nodes.push(root);
		int size = 1;
		int nextSize = 0;
		vector<int> result;
		TreeNode* node;
		int level = 1;
		while(!nodes.empty())
		{
			node = nodes.front();
			nodes.pop();
			result.push_back(node->val);
			if(node->left)
			{
				nodes.push(node->left);
				nextSize += 1;
			}
			if(node->right)
			{
				nodes.push(node->right);
				nextSize += 1;
			}
			size--;
			if(0 == size)
			{
				size = nextSize;
				nextSize = 0;
				vector<int> tempResult(result);
				//如果当前层数是奇数,需要逆置
				if((level & 1) == 0)
				{
					reverse(tempResult.begin() , tempResult.end());
				}
				level++;//层数要累加
				results.push_back(tempResult);
				result.clear();
			}
		}
		return results;
    }
};

void print(vector<int>& result)
{
	if(result.empty())
	{
		cout << "no result" << endl;
		return;
	}
	int size = result.size();
	for(int i = 0 ; i < size ; i++)
	{
		cout << result.at(i) << " " ;
	}
	cout << endl;
}

void process()
{
	 vector<int> nums;
	 int value;
	 int num;
	 Solution solution;
	 vector<int> result;
	 while(cin >> num )
	 {
		 nums.clear();
		 for(int i = 0 ; i < num ; i++)
		 {
			 cin >> value;
			 nums.push_back(value);
		 }
	 }
}

int main(int argc , char* argv[])
{
	process();
	getchar();
	return 0;
}