LeetCode 题解（32）: 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]
]

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}".

题解：

用一个bool变量表示正向或反向，用deque代替queue，因为deque可以正向遍历，也可反向遍历。每一层节点入queue后，在queue最后加入一个NULL节点。每次NULL节点出队列时，正向或反向扫描queue中当前的节点val，并保存在vector中，遍历完成后在queue中加入NULL节点。每次弹出NULL节点时，先判断队列是否为空，若为空，则说明当前NULL节点是最后一个节点，跳出循环，返回。

class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int>> result;
        
        if(!root)
            return result;
        
        vector<int> temp;
        temp.push_back(root->val);
        result.push_back(temp);
        bool reverse = false;
        deque<TreeNode*> nodeQ;
        nodeQ.push_back(root);
        nodeQ.push_back(NULL);
        
        reverse = !reverse;
        temp.erase(temp.begin(), temp.end());
        
        while(!nodeQ.empty())
        {
            TreeNode* node = nodeQ.front();
            nodeQ.pop_front();
            
            if(node)
            {
                if(node->left)
                    nodeQ.push_back(node->left);
                if(node->right)
                    nodeQ.push_back(node->right);
            }
            else
            {
                if(nodeQ.empty())
					break;
				if(reverse)
                {
                    for(deque<TreeNode*>::reverse_iterator rit = nodeQ.rbegin(); rit != nodeQ.rend(); rit++)
                    {
                        temp.push_back((*rit)->val);
                    }
                }
                else
                {
                    for(deque<TreeNode*>::iterator it = nodeQ.begin(); it != nodeQ.end(); it++)
                    {
                        temp.push_back((*it)->val);
                    }
                }
                reverse = !reverse;
                result.push_back(temp);
                temp.erase(temp.begin(), temp.end());
                nodeQ.push_back(NULL);
            }
        }
        return result;
    }
};