PTA-1020——Tree Traversals

Posted 2021-02-06 orangecyh

题目:

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 (≤), 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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

分析:

二叉树（根据后序、中序遍历还原二叉树）、BFS

参考博客：点击前往

代码:

 1 #include<iostream>
 2 #include<queue>
 3 using namespace std;
 4 struct Node{
 5     int data;
 6     Node* lchild;
 7     Node* rchild;
 8 };
 9 
10 int post[31];
11 int in[31];
12 int n;
13 
14 //根据后序和中序遍历创建二叉树 
15 Node* create(int postL,int postR,int inL,int inR){
16     if(postL>postR){
17         return NULL;
18     }
19     Node* root=new Node;
20     root->data=post[postR];        //后序的末尾是根 
21     int k;
22     for(k=inL;k<=inR;k++){        //中序中寻找和后序末尾值相同的结点 
23         if(in[k]==post[postR]){
24             break;
25         }
26     }
27     int numLeft=k-inL;
28     root->lchild=create(postL,postL+numLeft-1,inL,k-1);
29     root->rchild=create(postL+numLeft,postR-1,k+1,inR);
30     return root; 
31 }
32 
33 int num=0;
34 //宽搜读取二叉树 
35 void BFS(Node* root){
36     queue<Node*> q;
37     q.push(root);
38     while(!q.empty()){
39         num++;
40         Node* now=q.front();
41         cout<<now->data;
42         if(num<n){
43             cout<<" ";
44         }
45         q.pop();
46         if(now->lchild!=NULL){
47             q.push(now->lchild);
48         }
49         if(now->rchild!=NULL){
50             q.push(now->rchild);
51         }
52     }
53 }
54 
55 int main(){
56     cin>>n;
57     for(int i=0;i<n;i++){
58         cin>>post[i];
59     }
60     for(int i=0;i<n;i++){
61         cin>>in[i];
62     }
63     Node* root=create(0,n-1,0,n-1);
64     BFS(root);
65     return 0;
66 }