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;
    }
};