1020. Tree Traversals (25)
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:7 2 3 1 5 7 6 4 1 2 3 4 5 6 7Sample Output:
4 1 6 3 5 7 2
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<queue> 6 #include<vector> 7 #include<cmath> 8 using namespace std; 9 int post[35],in[35]; 10 struct node{ 11 int v; 12 node *l,*r; 13 node(){ 14 l=r=NULL; 15 } 16 }; 17 void BuildTree(int *post,int *in,int n,node *&h){ 18 if(!n){ 19 return; 20 } 21 h=new node(); 22 int mid=post[n-1]; 23 h->v=mid; 24 int i; 25 for(i=0;i<n;i++){ 26 if(in[i]==mid){ 27 break; 28 } 29 } 30 BuildTree(post,in,i,h->l); 31 BuildTree(post+i,in+i+1,n-1-i,h->r); 32 } 33 int main(){ 34 int n,i; 35 node *h; 36 scanf("%d",&n); 37 for(i=0;i<n;i++){ 38 scanf("%d",&post[i]); 39 } 40 for(i=0;i<n;i++){ 41 scanf("%d",&in[i]); 42 } 43 BuildTree(post,in,n,h); 44 node cur; 45 queue<node> q; 46 q.push(*h); 47 printf("%d",h->v); 48 while(!q.empty()){ 49 cur=q.front(); 50 q.pop(); 51 if(cur.l!=NULL){ 52 q.push(*(cur.l)); 53 printf(" %d",cur.l->v); 54 } 55 if(cur.r!=NULL){ 56 q.push(*(cur.r)); 57 printf(" %d",cur.r->v); 58 } 59 } 60 return 0; 61 }