标签:c++ pat
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<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define N 3005
struct node
{
int l,r;
}a[N];
int post[35],in[35];
int gettree(int lp,int rp,int li,int ri)
{
if(lp>=rp)return -1;
int i,x=post[rp-1];
for(i=li;i<ri;i++)
if(in[i]==x)break;
a[x].l=gettree(lp,rp-ri+i,li,i);
a[x].r=gettree(rp-ri+i,rp-1,i+1,ri);
return x;
}
int main()
{
int n,i,root,x,ans;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)scanf("%d",&post[i]);
for(i=0;i<n;i++)scanf("%d",&in[i]);
root=gettree(0,n,0,n);
queue<int>que;
que.push(root);
ans=0;
while(que.size())
{
x=que.front();
if(!ans)
{
printf("%d",x);
ans=1;
}
else
printf(" %d",x);
if(a[x].l!=-1)que.push(a[x].l);
if(a[x].r!=-1)que.push(a[x].r);
que.pop();
}
printf("\n");
}
return 0;
}
1020. Tree Traversals (25)
标签:c++ pat
原文地址:http://blog.csdn.net/jason_wang1989/article/details/44021095