LeetCode #103 - Binary Tree Zigzag Level Order Traversal

题目描述：

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]
]

将层序遍历的奇数层的向量元素反过来即可。

class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        queue<pair<TreeNode*,int> > q; 
		vector<pair<int,int> > v;
		if(root==NULL)
		{
		    vector<vector<int> > result;
		    return result;
		}
		else if(root!=NULL)
		{
			pair<TreeNode*,int> p;
			p.first=root;
			p.second=0;
			q.push(p);
			while(!q.empty())
			{
				p=q.front();
				pair<int,int> x;
				x.first=p.first->val;
				x.second=p.second;
				v.push_back(x);
				q.pop();
				if(p.first->left!=NULL)
				{
					pair<TreeNode*,int> l;
					l.first=p.first->left;
					l.second=p.second+1;
					q.push(l);
				}
				if(p.first->right!=NULL)
				{
					pair<TreeNode*,int> r;
					r.first=p.first->right;
					r.second=p.second+1;
					q.push(r);
				}
			}
		}
		int n=v.size();
		int m=v[n-1].second;
		vector<vector<int> > result(m+1);
		for(int i=0;i<n;i++)
		{
			result[v[i].second].push_back(v[i].first);
		}
		vector<vector<int> > Result=result;
		for(int i=0;i<=m;i++)
		{
		    if(i%2==1)
		    {
		        for(int j=0;j<result[i].size();j++)
		        {
		            Result[i][j]=result[i][result[i].size()-1-j];
		        }
		    }
		}
		return Result;
    }
};