1020. Tree Traversals (序列建树)

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;

}