Source:

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:

812 11 20 17 1 15 8 512 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

Keys:

• 二叉树的建立
• 二叉树的遍历

Attention:

• 开始方向弄反了-,-

Code:

1 /* 2 Data: 2019-05-31 21:17:07 3 Problem: PAT_A1127#ZigZagging on a Tree 4 AC: 33:50 5  6 题目大意： 7 假设树的键值为不同的正整数； 8 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； 9 */10 11 #include
     
      12 #include
      
       13 #include
       
        14 using namespace std;15 const int M=35;16 int in[M],post[M];17 struct node18 {19     int data,layer;20     node *lchild,*rchild;21 };22 23 node *Create(int postL, int postR, int inL, int inR)24 {25     if(postL > postR)26         return NULL;27     node *root = new node;28     root->data = post[postR];29     int k;30     for(k=inL; k<=inR; k++)31         if(in[k]==root->data)32             break;33     int numLeft = k-inL;34     root->lchild = Create(postL, postL+numLeft-1, inL,k-1);35     root->rchild = Create(postL+numLeft, postR-1, k+1,inR);36     return root;37 }38 39 void Travel(node *root)40 {41     queue
        
          q;42     stack
         
           s;43     root->layer=1;44     q.push(root);45     while(!q.empty())46     {47         root = q.front();q.pop();48         if(root->layer%2==0)49         {50             while(!s.empty())51             {52                 printf(" %d", s.top()->data);53                 s.pop();54             }55             printf(" %d", root->data);56         }57         else{58             if(root->layer==1)59                 printf("%d", root->data);60             else61                 s.push(root);62         }63         if(root->lchild){64             root->lchild->layer=root->layer+1;65             q.push(root->lchild);66         }67         if(root->rchild){68             root->rchild->layer=root->layer+1;69             q.push(root->rchild);70         }71     }72     while(!s.empty())73     {74         printf(" %d", s.top()->data);75         s.pop();76     }77 }78 79 int main()80 {81 #ifdef    ONLINE_JUDGE82 #else83     freopen("Test.txt", "r", stdin);84 #endif85 86     int n;87     scanf("%d", &n);88     for(int i=0; i