A1020. Tree Traversals

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:

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

Sample Output:

4 1 6 3 5 7 2

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 using namespace std;
 5 typedef struct NODE{
 6     struct NODE* lchild, *rchild;
 7     int data;
 8 }node;
 9 int in[31], post[31];
10 int N;
11 node* create(int inL, int inR, int postL, int postR){
12     if(postL > postR){
13         return NULL;
14     }
15     int head = postR;
16     node * root = new node;
17     root->data = post[head];
18     int i;
19     for(i = inL; i <= inR; i++){
20         if(in[i] == post[head])
21             break;
22     }
23     int numL = i - inL;
24     root->lchild = create(inL, i - 1, postL, postL + numL - 1);
25     root->rchild = create(i + 1, inR, postL + numL, postR - 1);
26     return root;
27 }
28 void levelOrder(node* tree){
29     queue<node*> Q;
30     Q.push(tree);
31     node* temp;
32     int cnt = 0;
33     while(Q.empty() == false){
34         temp = Q.front();
35         cnt++;
36         Q.pop();
37         printf("%d", temp->data);
38         if(cnt != N)
39             printf(" ");
40         if(temp->lchild != NULL)
41             Q.push(temp->lchild);
42         if(temp->rchild != NULL)
43             Q.push(temp->rchild);
44     }
45 }
46 int main(){
47     scanf("%d", &N);
48     for(int i = 0; i < N; i++)
49         scanf("%d", &post[i]);
50     for(int i = 0; i < N; i++)
51         scanf("%d", &in[i]);
52     node* tree = create(0, N - 1, 0, N - 1);
53     levelOrder(tree);
54     cin >> N;
55     return 0;
56 }
View Code

 

总结：

1、题意：给出后序遍历和中序遍历，求层序遍历。

2、层序遍历时不要忘记在访问完元素后pop。

3、创建二叉树结束的条件是后序遍历的区间为空（postL > postR）。

4、注意控制最后一个空格不要输出，可以设置一个计数器，为N时不输出空格。

posted @ 2018-02-07 15:08  ZHUQW  阅读(142)  评论(0)  编辑  收藏  举报