LeetCode(103)Binary Tree Zigzag Level Order Traversal

题目如下:

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?

题目分析:

在前面1题的基础上，本题增加了一个小变化，就是输出的时候，一行从左到右输出，另一行从右到左输出。于是可以设定flag来代表奇偶的变化。
代码如下:

//36ms过大集合/** * 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>> res_vec_vec;        if(root==NULL)            return res_vec_vec;        queue<queue<TreeNode*>> que_que;        queue<TreeNode*> que;        que.push(root);        que_que.push(que);        int flag=1;        while(!que_que.empty()){            queue<TreeNode*> tmp_src_queue=que_que.front();            queue<TreeNode*> tmp_out_queue;            vector<int> tmp_vec;            while(!tmp_src_queue.empty()) {                tmp_vec.push_back(tmp_src_queue.front()->val);                TreeNode* tmp_node=tmp_src_queue.front();                if(tmp_node->left!=NULL)                    tmp_out_queue.push(tmp_node->left);                if(tmp_node->right!=NULL)                    tmp_out_queue.push(tmp_node->right);                tmp_src_queue.pop();            }            if(flag>0) {                res_vec_vec.push_back(tmp_vec);            }else{                int k=0;                int h=tmp_vec.size()-1;                while(k<h){                    int tmp=tmp_vec[k];                    tmp_vec[k]=tmp_vec[h];                    tmp_vec[h]=tmp;                    k++;                    h--;                }                res_vec_vec.push_back(tmp_vec);            }            flag=0-flag;            if(!tmp_out_queue.empty())//如果没有这个条件判断，程序就死循环了                que_que.push(tmp_out_queue);            que_que.pop();        }        return res_vec_vec;    }};

update: 2014-12-03

/** * 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* current;        queue<TreeNode*> que;        queue<TreeNode*> temp_que;        vector<int> vec;        vector<vector<int> > nested_vec;        if(root != NULL) que.push(root);        bool is_odd = true;        while(!que.empty()) {            current = que.front();            if(is_odd) {                vec.push_back(current->val);            }else {                vec.insert(vec.begin(),current->val); //NOTE: insert(参数1,参数2), 其中参数1不能用数字表示位置 ×: vec.insert(0,current->val);            }            que.pop();            if (current->left != NULL)                temp_que.push(current->left);            if (current->right != NULL)                temp_que.push(current->right);                if (que.size()==0) {                is_odd = 1 - is_odd;                que = temp_que;                nested_vec.push_back(vec);                vec.clear();                queue<TreeNode*> q;                temp_que.swap(q);//NOTE: std::queue没有clear()函数,可以swap一个空queue相当于清零。             }        }        return nested_vec;    }};

update: 2015-03-24

1. 做了一个小改进，参照前一题目，使用size来表示每层的节点个数，简化代码。其它地方基本差不多。

//10ms/** * 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) {        queue<TreeNode*> q;        if (root != NULL) {            q.push(root);        }        vector<int> inner;        vector<vector<int> > outer;        TreeNode* node;        bool isReversed = false;        while(!q.empty()) {            int size = q.size();            inner.clear();            while (size-- > 0) {                node = q.front();                q.pop();                inner.push_back(node->val);                if (node->left != NULL)                    q.push(node->left);                if (node->right != NULL)                    q.push(node->right);            }            if(isReversed) {                reverse(inner.begin(), inner.end());            }            outer.push_back(inner);            isReversed = 1 - isReversed;        }        return outer;    }};