LeetCode_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]
]

　　和level order traversal 唯一的不同是，偶数行逆序输出，逆序可是使用stack来实现。其他的实现和level order实现大同小异

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> result ;
        if(root ==NULL) return result;
        std::queue<TreeNode *> q,p;
        std::stack<int> kit;
        vector<int> mResult;
        bool f = false ;
        q.push(root);
        while(!q.empty()){
          
           mResult.clear();
           if(f)
           {
            while(!q.empty()){            
                root = q.front(); q.pop();
                kit.push(root->val);
                if(root->left)
                    p.push(root->left);
                if(root->right)
                    p.push(root->right);
            }
            while(!kit.empty())
            {
              mResult.push_back(kit.top());
              kit.pop();
            }
            f = false;
           }else{
           
              while(!q.empty()){            
                root = q.front(); q.pop();
                mResult.push_back(root->val);
                if(root->left)
                    p.push(root->left);
                if(root->right)
                    p.push(root->right);
              }
              f = true ;
           }
           result.push_back(mResult);
            while(!p.empty()){
            q.push(p.front());
            p.pop();            
           }
           
        }
        
        return result;
        
    }
};