[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;    }};