PAT1020

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1020

题目较简单，根据后序和中序遍历写出二叉树的层序遍历，我的思想：根据2个序列递归构建二叉树，再层序遍历。

只要计算好下标就好了。

 

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 using namespace std;
 5 
 6 struct Node
 7 {
 8     int val;
 9     Node* left;
10     Node* right;
11 };
12 
13 /*递归建树*/
14 Node*  buildtree(vector<int> post, int i, int j, vector<int> in, int k, int h)
15 {
16     if(i > j || k > h)
17         return NULL;
18     Node* n = new Node;
19     n->val = post[j];
20     int mid(-1);
21     for(int i=k; i<=h; ++i)
22         if(in[i] == post[j])
23         {
24             mid = i;
25             break;
26         }
27     n->left = buildtree(post, i, i+mid-k-1, in, k, mid-1);
28     n->right = buildtree(post, i+mid-k, j-1, in, mid+1, h);
29     return n;
30 }
31 
32 int main()
33 {
34     int n;
35     while(cin>>n)
36     {
37         vector<int> post(n, 0);
38         vector<int> in(n, 0);
39         for(int i=0; i<n; ++i)
40             cin>>post[i];
41         for(int i=0; i<n; ++i)
42             cin>>in[i];
43         Node *root = buildtree(post, 0, n-1, in, 0, n-1);
44         queue<Node*> q;
45         q.push(root);
46         int m(0);
47         while(!q.empty())
48         {
49             if(q.front()->left != NULL)
50                 q.push(q.front()->left);
51             if(q.front()->right != NULL)
52                 q.push(q.front()->right);
53             if(m == n-1)
54                 cout<<q.front()->val<<endl;
55             else
56                 cout<<q.front()->val<<" ";
57             q.pop();
58             ++m;
59         }
60     }
61     return 0;
62 }

 

posted @ 2013-10-08 19:51  coding_monkey  阅读(154)  评论(0)  编辑  收藏  举报