PAT_A1127#ZigZagging on a Tree

Source:

PAT A1127 ZigZagging on a Tree (30 分)

Description:

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:

1 11 5 8 17 12 20 15

Keys:

• 二叉树的建立和遍历

Attention:

• 开始方向弄反了-,-

Code:

 /*
 Data: 2019-05-31 21:17:07
 Problem: PAT_A1127#ZigZagging on a Tree
 AC: 33:50

 题目大意：
 假设树的键值为不同的正整数；
 给出二叉树的中序和后序遍历，输出二叉树的“之字形”层次遍历；

 基本思路：
 建树，层次遍历，依次用队列和堆存储结点值，输出即可
 */

 #include<cstdio>
 #include<stack>
 #include<queue>
 using namespace std;
 const int M=;
 int in[M],post[M];
 struct node
 {
     int data,layer;
     node *lchild,*rchild;
 };

 node *Create(int postL, int postR, int inL, int inR)
 {
     if(postL > postR)
         return NULL;
     node *root = new node;
     root->data = post[postR];
     int k;
     for(k=inL; k<=inR; k++)
         if(in[k]==root->data)
             break;
     int numLeft = k-inL;
     root->lchild = Create(postL, postL+numLeft-, inL,k-);
     root->rchild = Create(postL+numLeft, postR-, k+,inR);
     return root;
 }

 void Travel(node *root)
 {
     queue<node*> q;
     stack<node*> s;
     root->layer=;
     q.push(root);
     while(!q.empty())
     {
         root = q.front();q.pop();
         if(root->layer%==)
         {
             while(!s.empty())
             {
                 printf(" %d", s.top()->data);
                 s.pop();
             }
             printf(" %d", root->data);
         }
         else{
             if(root->layer==)
                 printf("%d", root->data);
             else
                 s.push(root);
         }
         if(root->lchild){
             root->lchild->layer=root->layer+;
             q.push(root->lchild);
         }
         if(root->rchild){
             root->rchild->layer=root->layer+;
             q.push(root->rchild);
         }
     }
     while(!s.empty())
     {
         printf(" %d", s.top()->data);
         s.pop();
     }
 }

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int n;
     scanf("%d", &n);
     for(int i=; i<n; i++)
         scanf("%d", &in[i]);
     for(int i=; i<n; i++)
         scanf("%d", &post[i]);
     node *root = Create(,n-,,n-);
     Travel(root);

     return ;
 }