Binary Tree Zigzag Level Order Traversal

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}".

Solution:

/** * Definition for a binary tree node. * 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> > res;        bfs(root, res);                return res;    }        void bfs(TreeNode *root, vector<vector<int> > &res)    {        if(!root) return ;        queue<TreeNode *> q;        q.push(root);        int count = 1, level = 1;        vector<int> v;        while(!q.empty())        {            TreeNode *tn = q.front();            q.pop();            v.push_back(tn->val);            if(tn->left) q.push(tn->left);            if(tn->right) q.push(tn->right);            if(!--count)            {                count = q.size();                if(level % 2 == 0) reverse(v.begin(), v.end());                res.push_back(v);                ++level;                v.clear();            }        }    }};