PAT A1127 ZigZagging on a Tree （30 分）——二叉树，建树，层序遍历

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <queue>
 4 using namespace std;
 5 const int maxn=31;
 6 int pos[maxn],in[maxn];
 7 int n,m;
 8 vector<int> res[31];
 9 struct node{
10     int data;
11     node* left,*right;
12     int lvl;
13 };
14 node* create(int inl,int inr,int posl,int posr){
15     if(inl > inr) return NULL;
16     node* root = new node;
17     root->data = pos[posr];
18     int k;
19     for(k=inl;k<=inr;k++){
20         if(pos[posr]==in[k]) break;
21     }
22     int numleft=k-inl;
23     root->left = create(inl,k-1,posl,posl+numleft-1);
24     root->right = create(k+1,inr,posl+numleft,posr-1);
25     return root;
26 }
27 void pr(node* root){
28     queue<node*> q;
29     root->lvl=0;
30     q.push(root);
31     while(!q.empty()){
32         node* now=q.front();
33         q.pop();
34         res[now->lvl].push_back(now->data);
35         if(now->left!=NULL){
36             now->left->lvl=now->lvl+1;
37             q.push(now->left);
38         }
39         if(now->right!=NULL){
40             now->right->lvl=now->lvl+1;
41             q.push(now->right);
42         }
43     }
44 }
45 int main(){
46     scanf("%d",&n);
47     for(int i=0;i<n;i++){
48         scanf("%d",&in[i]);    
49     }
50     for(int i=0;i<n;i++){
51         scanf("%d",&pos[i]);    
52     }
53     node* root = create(0,n-1,0,n-1);
54     pr(root);
55     int cnt=0;
56     for(int i=0;i<32;i++){
57         if(i%2==0){
58             for(int j=res[i].size()-1;j>=0;j--){
59                 printf("%d",res[i][j]);
60                 cnt++;
61                 if(cnt<n)printf(" ");
62             }
63         }
64         else{
65             for(int j=0;j<res[i].size();j++){
66                 printf("%d",res[i][j]);
67                 cnt++;
68                 if(cnt<n)printf(" ");
69             }
70         }
71         if(cnt==n)break;
72     }
73 }

View Code

注意点：最简单的一道30分题了。只要会建树，再层序遍历，把每层的数据保存起来，再根据层数正着输出或是反向输出就ac了。

ps:也可以用deque双向队列更方便的实现。每层的输出其实可以不用开vector数组，一层遍历完后输出就好了。

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