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;

        

    }

};