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]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：

类似http://blog.csdn.net/lanxu_yy/article/details/11820189。只是在输出最终结果前需要将部分level反序。

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 {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        TreeNode * q1[10000];        bool q2[10000];        int begin=0;        int end=0;                if(root == NULL)        {            vector<vector<int> > v;            return v;        }        else        {            vector<vector<int> > v;            q1[end] = root;            q2[end++] = true;                        bool level = true;            bool cur = true;                        vector<int> * tmp = new vector<int>();            while(begin!=end)            {                TreeNode * p = q1[begin];                cur = q2[begin++];                                if(cur != level)                {                    if(!level)                    {                        for(int k=0;k<tmp->size()/2;k++)                        {                            int t = tmp->at(k);                            tmp->at(k) = tmp->at(tmp->size()-1-k);                            tmp->at(tmp->size()-1-k) = t;                        }                        v.push_back(*tmp);                    }                    else                    {                        v.push_back(*tmp);                    }                    delete tmp;                    tmp = new vector<int>();                    level = !level;                                    }                if(cur == level)                {                    if(p->left != NULL)                    {                        q1[end] = p->left;                        q2[end++] = !cur;                    }                    if(p->right != NULL)                    {                        q1[end] = p->right;                        q2[end++] = !cur;                    }                    tmp->push_back(p->val);                }                            }            if(!level)            {                for(int k=0;k<tmp->size()/2;k++)                {                    int t = tmp->at(k);                    tmp->at(k) = tmp->at(tmp->size()-1-k);                    tmp->at(tmp->size()-1-k) = t;                }                v.push_back(*tmp);            }            else            {                                v.push_back(*tmp);            }            return v;        }    }};