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.
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.
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.
8 12 11 20 17 1 15 8 5 12 20 17 11 15 8 5 1
1 11 5 8 17 12 20 15
result:
1 #include<cstdio> 2 #include<vector> 3 #include<algorithm> 4 using namespace std; 5 const int maxn = 33; 6 int n,postOrder[maxn],inOrder[maxn]; 7 vector<vector<int> > ans(maxn); 8 int level1=0; 9 10 void createTree(int inL,int inR,int postL,int postR,int level2){ 11 if(postR<postL || inR<inL) return ; 12 int data = postOrder[postR]; 13 ans[level2].push_back(data); 14 // printf("%d ",data); 15 level1 = max(level1,level2); 16 int num=0; 17 while(inOrder[inL+num] != postOrder[postR]){ 18 num++; 19 } 20 // 这样生成的num指的是有num个lchild; 21 createTree(inL,inL+num-1,postL,postL +num-1,level2+1); 22 createTree(inL+num+1,inR,postL +num,postR - 1,level2+1); 23 return ; 24 } 25 26 27 int main(){ 28 scanf("%d",&n); 29 for(int i=0;i<n;i++){ 30 scanf("%d",&inOrder[i]); 31 } 32 for(int i=0;i<n;i++){ 33 scanf("%d",&postOrder[i]); 34 } 35 createTree(0,n-1,0,n-1,0); 36 int s = 0; 37 bool flag = false; 38 for(vector<vector<int> >::iterator i=ans.begin();i!=ans.end();++i){ 39 if(flag){ 40 for(vector<int>::iterator j=(*i).begin();j !=(*i).end();++j){ 41 printf("%d",(*j)); 42 s++; 43 if(s<n) printf(" "); 44 } 45 flag=false; 46 }else{ 47 reverse((*i).begin(),(*i).end()); 48 for(vector<int>::iterator j=(*i).begin(); j !=(*i).end();++j){ 49 printf("%d",(*j)); 50 s++; 51 if(s<n) printf(" "); 52 } 53 flag=true; 54 } 55 } 56 }
PAT A1127 ZigZagging on a Tree （30 分）
原文：https://www.cnblogs.com/bobyin/p/10495084.html
