LeetCode Binary Tree Zigzag Level Order Traversal

题目

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]
]

 

Z字形遍历。

实际就是第1层从左向右遍历，第二层从右向左遍历……

依然可以用bfs解决，

每扫完一层后将队列反转，

如果是奇数层，左节点先入队列；

如果是偶数层，右节点先入队列。

 

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
	struct pt	//bfs用点
	{
		TreeNode *node;	//探测到的位置
		int level;	//相应的层
		pt(TreeNode *n=NULL,int l=0):node(n),level(l){}
	};
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int>> ret;
		deque<pt> queue(1,pt(root));	//队列
		while(!queue.empty())	//bfs
		{	
			if(queue.front().node!=NULL)
			{
				if(queue.front().level>=ret.size())	//每换一层时将队列翻转
				{
					ret.push_back(vector<int>());
					reverse(queue.begin(),queue.end());
					continue;
				}
				ret[queue.front().level].push_back(queue.front().node->val);
				if(ret.size()%2==1)	//正向入队
				{
					queue.push_back(pt(queue.front().node->left,queue.front().level+1));
					queue.push_back(pt(queue.front().node->right,queue.front().level+1));
				}
				else	//反向入队
				{
					queue.push_back(pt(queue.front().node->right,queue.front().level+1));
					queue.push_back(pt(queue.front().node->left,queue.front().level+1));
				}
			}
			queue.pop_front();
		}
		return ret;
    }
};