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 7
Sample Output:
4 1 6 3 5 7 2
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
int post[],in[];
struct node{
int v;
node *l,*r;
node(){
l=r=NULL;
}
};
void BuildTree(int *post,int *in,int n,node *&h){
if(!n){
return;
}
h=new node();
int mid=post[n-];
h->v=mid;
int i;
for(i=;i<n;i++){
if(in[i]==mid){
break;
}
}
BuildTree(post,in,i,h->l);
BuildTree(post+i,in+i+,n--i,h->r);
}
int main(){
int n,i;
node *h;
scanf("%d",&n);
for(i=;i<n;i++){
scanf("%d",&post[i]);
}
for(i=;i<n;i++){
scanf("%d",&in[i]);
}
BuildTree(post,in,n,h);
node cur;
queue<node> q;
q.push(*h);
printf("%d",h->v);
while(!q.empty()){
cur=q.front();
q.pop();
if(cur.l!=NULL){
q.push(*(cur.l));
printf(" %d",cur.l->v);
}
if(cur.r!=NULL){
q.push(*(cur.r));
printf(" %d",cur.r->v);
}
}
return ;
}