[C++]LeetCode: 101 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.

思路：这道题的解法和思路和Binary
Tree Level Order Traversal差不多，只不过我们需要设个标志位来区分，这行是正序还是反序输出数组。我们利用队列来存储树的节点，并用NULL来区分每一行，记得当队列非空时，处理完每一行，都需要清空level数组，并加入NULL，区别下一行。我们设置tag, tag为True，表示正序，从左到右输出数组，tag为false, 表示反序，从右往左输出数组。

Attention:

1. 记得处理完每一行后，当队列非空时，清空数组level, 并且往队列中加入NULL，区分下一行。

2. reverse的用法

void reverse (BidirectionalIterator first, BidirectionalIterator last);

std::reverse(myvector.begin(),myvector.end()); 

3. 用tag来区分输出顺序。记得每次都需要tag取反，下次是反方向。

                if(tag)
                {
                    ret.push_back(level);
                }
                else
                {
                    reverse(level.begin(), level.end());
                    ret.push_back(level);
                }
                tag = !tag;

复杂度：O(N)，N为树的节点个数。

AC Code:

/**
 * Definition for binary tree
 * 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>> ret;
        if(root == NULL) return ret;
        vector<int> level;
        bool tag = true;

        queue<TreeNode*> TreeQ;
        TreeQ.push(root);
        TreeQ.push(NULL); //区分分层

        while(!TreeQ.empty())
        {
            TreeNode* node = TreeQ.front();
            TreeQ.pop();
            if(node != NULL)
            {
                level.push_back(node->val);
                if(node->left) TreeQ.push(node->left);
                if(node->right) TreeQ.push(node->right);
            }
            else
            {
                if(tag)
                {
                    ret.push_back(level);
                }
                else
                {
                    reverse(level.begin(), level.end());
                    ret.push_back(level);
                }
                tag = !tag;
                if(!TreeQ.empty())
                {
                    level.clear();
                    TreeQ.push(NULL);
                }
            }
        }

        return ret;
    }
};