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
见之前一篇日志《序列建树（递归） 》
#include <iostream>
using namespace std;
struct LNode
{
LNode *lchild,*rchild;
int data;
};
int in[31];
int pos[31];
LNode* fun(int pos[],int f1,int r1,int in[],int f2,int r2)//生成树
{
if(f1>r1) return NULL;
LNode *p=(LNode*)malloc(sizeof(LNode));
p->lchild=NULL;
p->rchild=NULL;
p->data=pos[r1];
int i;
for(i=f2;i<=r2;i++)
if(in[i]==pos[r1]) break;
p->lchild=fun(pos,f1,f1+i-f2-1,in,f2,i-1);
p->rchild=fun(pos,f1+i-f2,r1-1,in,i+1,r2);
return p;
}
int main()
{
int n;
while(cin>>n)
{
int i;
for(i=0;i<n;i++)
cin>>pos[i];
for(i=0;i<n;i++)
cin>>in[i];
LNode *q=(LNode*)malloc(sizeof(LNode));
q=fun(pos,0,n-1,in,0,n-1);
LNode* que[40]; //层次遍历
int front=0;int rear=0;
rear++;
que[rear]=q;
bool fir=true;
while(front!=rear)
{
front++;
if(fir)
{cout<<que[front]->data;fir=false;}
else cout<<" "<<que[front]->data;
if(que[front]->lchild!=NULL)
que[++rear]=que[front]->lchild;
if(que[front]->rchild!=NULL)
que[++rear]=que[front]->rchild;
}
cout<<endl;
}
return 0;
}