1127 ZigZagging on a Tree （30 分）

1127 ZigZagging on a Tree （30 分）

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

题意：中序和后序建树，然后按zigzagging order输出。

分析：层序遍历的时候将节点输出到容器中，最后输出的时候根据奇数还是偶数来输出结点

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-28-14.24.50
 * Description : A1127
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 #include<queue>
 using namespace std;
 ;
 int n,post[maxn],in[maxn];
 vector<int> ans[maxn];
 struct Node{
     int val,layer;
     Node* lchild;
     Node* rchild;
     Node(int _val,int _layer){
         val=_val;
         lchild=NULL;
         rchild=NULL;
         layer=_layer;
     }
 };
 ;
 Node* create(int inL,int inR,int postL,int postR,int layer){
     if(inL>inR) return NULL;
     if(layer>maxlayer) maxlayer=layer;
     int rootVal=post[postR];
     Node* root=new Node(rootVal,layer);
     int k;
     for(int i=inL;i<=inR;i++){
         if(post[postR]==in[i]){
             k=i;
             break;
         }
     }
     int numLeft=k-inL;
     root->lchild=create(inL,k-,postL,postL+numLeft-,layer+);
     root->rchild=create(k+,inR,postL+numLeft,postR-,layer+);
     return root;
 }
 
 void BFS(Node* root){
     queue<Node*> q;
     q.push(root);
     while(!q.empty()){
         Node* now=q.front();
         q.pop();
         ans[now->layer].push_back(now->val);
         if(now->lchild!=NULL) q.push(now->lchild);
         if(now->rchild!=NULL) q.push(now->rchild);
     }
 }
 
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     scanf("%d",&n);
     ;i<n;i++){
         scanf("%d",&in[i]);
     }
     ;i<n;i++){
         scanf("%d",&post[i]);
     }
     Node* root=create(,n-,,n-,);
     BFS(root);
     ;i<=maxlayer;i++){
         ){
             printf(]);
             continue;
         }
         ==){
             ;j<ans[i].size();j++){
                 printf(" %d",ans[i][j]);
             }
         }
         else{
             ;j>=;j--){
                 printf(" %d",ans[i][j]);
             }
         }
     }
     ;
 }