剑指offer 面试题. 按之字形顺序打印二叉树

题目描述

请实现一个函数按照之字形打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右至左的顺序打印,第三行按照从左到右的顺序打印,其他行以此类推。
 

方法1:

正常层次遍历,利用普通队列。逢奇数行(从0算起)就把该层结果逆序。

 1 class Solution {
 2 public:
 3     vector<vector<int> > Print(TreeNode* pRoot) {
 4         if(pRoot==nullptr){return {};}
 5         queue<TreeNode*> my_queue;
 6         my_queue.push(pRoot);
 7         vector<vector<int>>res;
 8         vector<int> vec;
 9         int cnt=0;
10         while(not my_queue.empty()){
11             vec.clear();
12             int cur_siz=my_queue.size();
13             for(int i=0;i<cur_siz;++i){
14                 auto cur=my_queue.front();
15                 my_queue.pop();
16                 vec.push_back(cur->val);
17                 if(cur->left){
18                     my_queue.push(cur->left);
19                 }
20                 if(cur->right){
21                     my_queue.push(cur->right);
22                 }
23             }
24             if(cnt&1){
25                 reverse(vec.begin(),vec.end());
26             }
27             res.push_back(vec);
28             ++cnt;
29         }
30         return res;
31     }
32 };

 

方法2:

利用双端队列,队列中顺序始终是正常的,但奇数行从右向左遍历,pop右边的,孩子push进左边。偶数行从左向右遍历,pop左边的,孩子push进右边。

 1 class Solution {
 2 public:
 3     vector<vector<int> > Print(TreeNode* pRoot) {
 4         if(pRoot==nullptr){return {};}
 5         deque<TreeNode*> my_queue;
 6         my_queue.push_back(pRoot);
 7         vector<vector<int>>res;
 8         int cnt=0;
 9         while(not my_queue.empty()){
10             res.push_back({});
11             int cur_siz=my_queue.size();
12             decltype(pRoot) cur;
13             for(int i=0;i<cur_siz;++i){
14                 if(cnt&1){
15                     cur=my_queue.back();
16                     my_queue.pop_back();
17                     if(cur->right){
18                         my_queue.push_front(cur->right);
19                     }
20                     if(cur->left){
21                         my_queue.push_front(cur->left);
22                     }
23                 }
24                 else{
25                     cur=my_queue.front();
26                     my_queue.pop_front();
27                     if(cur->left){
28                         my_queue.push_back(cur->left);
29                     }
30                     if(cur->right){
31                         my_queue.push_back(cur->right);
32                     }
33                 }
34                 res.back().push_back(cur->val);
35             }
36             ++cnt;
37         }
38         return res;
39     }
40 };

 

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转载自www.cnblogs.com/FdWzy/p/12306131.html