PAT甲级 1020. Tree Traversals (25)

题目：

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

思路：

1.根据后序和中序的序列构建二叉树

2.利用广度优先搜索

注意：

曾经利用深度优先搜索来做，但第五个测试点出现段错误，说明应采用广度优先搜索

代码：

#include<iostream>#include<vector>using namespace std;#ifndef Maxnum#define Maxnum 31struct node{int left;int right;};node tree[Maxnum];int postor[Maxnum];int inor[Maxnum];//构建二叉树int creattree(int post_be,int post_end,int in_be,int in_end){if(post_be<post_end){int root=postor[post_end];int i;for(i=0;i<Maxnum;++i){if(inor[i]==root)   break;}int len,left,right;//左子树len=i-in_be;tree[root].left=creattree(post_be,post_be+len-1,in_be,i-1);//右子树len=in_end-i;tree[root].right=creattree(post_end-len,post_end-1,i+1,in_end);return root;}else if(post_be==post_end){return postor[post_end];}elsereturn 0;}//广度优先搜索vector<int> q;vector<int> q1;void bfs(int root){if(root!=0){q1.push_back(root);//插入根节点q.push_back(root);while(!q1.empty()){int p=q1.front();q1.erase(q1.begin());if(tree[p].left!=0){q1.push_back(tree[p].left);q.push_back(tree[p].left);}if(tree[p].right!=0){q1.push_back(tree[p].right);q.push_back(tree[p].right);}}}}int main(){int N;cin>>N;int i;for(i=0;i<N;++i){cin>>postor[i];}for(i=0;i<N;++i){cin>>inor[i];}int root;root=creattree(0,N-1,0,N-1);//广度优先搜索bfs(root);for(i=0;i<N;++i){cout<<q[i];if(i!=N-1)cout<<" ";elsecout<<endl;}system("pause");return 0;}#endif