PAT 1127 ZigZagging on a Tree (30)

发布时间：2020-12-14 03:23:47 所属栏目：大数据来源：网络整理

导读：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 lev

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 (<= 30),the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

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:




题目大意：通过中序遍历和后序遍历构建一棵树， 并交替的输出层序遍历；
思路：dfs()构建这棵树， levelOrder()交替的层序遍历这棵树；
　　 用两个二维数组来记录层序遍历， level1记录从右到左边的遍历， level2用来记录从左到右的遍历
    设置两个计数cnt1，cnt2来记录level1和了level2的位置
注意点：应该对建立树的数组进行初始化，初始化为负数， 若不初始化， 默认值是1， 然而数组有下标是0，会导致第一位数不能遍历，导致错误1 11 5 8 17 12 20 15

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 using namespace std;
 5 vector<int> in(30),post(30),level1[30],level2[30];
 6 int v[30][2],root;
 7 void dfs(int &index,int inl,int inr,int postl,int postr){
 8   int temp=post[postr],i=inl;
 9   if(inl>inr) return;
10   index = postr;
11   while(i<=inr && in[i]!=temp) i++;
12   dfs(v[index][0],inl,i-1,postl,postl+(i-inl)-1);
13   dfs(v[index][1],i+1,inr,postl+i-inl,postr-1);
14 }
15 
16 void levelOrder(){
17   queue<int> q1,q2;
18   q1.push(root);
19   int cnt1=0,cnt2=0;
20   while(q1.size() || q2.size()){
21       int temp,flag1=0,flag2=0;
22       while(q1.size()){
23         temp=q1.front();
24         q1.pop();
25         level1[cnt1].push_back(post[temp]);
26         if(v[temp][0]!=-1) q2.push(v[temp][0]);
27         if(v[temp][1]!=-1) q2.push(v[temp][1]);
28         flag1=1;
29       }
30       if(flag1) cnt1++;
31       while(q2.size()){
32         temp=q2.front();
33         q2.pop();
34         level2[cnt2].push_back(post[temp]);
35         if(v[temp][0]!=-1) q1.push(v[temp][0]);
36         if(v[temp][1]!=-1) q1.push(v[temp][1]);
37         flag2=1;
38       }
39       if(flag2) cnt2++;
40   }
41 }
42 int main(){
43   int n,i,j;
44   cin>>n;
45   fill(v[0],v[0]+60,-1);
46   for(i=0; i<n; i++) cin>>in[i];
47   for(i=0; i<n; i++) cin>>post[i];
48   dfs(root,0,n-1,n-1);
49   levelOrder();
50   cout<<level1[0][0];
51   for(i=1; i<n; i++)
52       if(i%2==1) for(j=0; j<level2[i/2].size(); j++) cout<<" "<<level2[i/2][j];
53       else for(j=level1[i/2].size()-1; j>=0; j--) cout<<" "<<level1[i/2][j];
54   return 0;
55 }

（编辑：李大同）

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