Tree Traversals (25)

题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

72 3 1 5 7 6 41 2 3 4 5 6 7

输出例子:

4 1 6 3 5 7 2

我的代码：

#include<iostream>#include<queue>using namespace std;struct Node{    int data;    Node *lchild,*rchild;};Node *creat(int h[],int z[],int n){    Node *t;int i;    if(n==0) return NULL;    t=new Node;    t->data=h[n-1];    for(i=0;i<n;i++)    {        if(z[i]==h[n-1]) break;    }    t->lchild=creat(h,z,i);    t->rchild=creat(h+i,z+i+1,n-i-1);    return t;}void ceng(Node *t){    queue<Node*>q;    int a[101],i,k=0;    if(t)    {        q.push(t);        while(!q.empty())        {            t=q.front();            q.pop();            a[k++]=t->data;            if(t->lchild) q.push(t->lchild);            if(t->rchild) q.push(t->rchild);        }    }    for(i=0;i<k;i++){if(i==0) cout<<a[i];else cout<<" "<<a[i];}}int main(){    int n,i,b[101],c[101];    cin>>n;    for(i=0;i<n;i++) cin>>b[i];    for(i=0;i<n;i++) cin>>c[i];    Node *r=creat(b,c,n);    ceng(r);    return 0;}