pat 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

 已知二叉树后序和中序遍历，求层序遍历。

重建二叉树后，层序遍历就bfs二叉树就可以了。

#include<iostream>
using namespace std;
#include<string.h>
#include<stdio.h>
#include<queue>
#define MS(a,b) memset(a,b,sizeof(a))
typedef struct node
{
int data;
node *left,*right;
}*Tree;
Tree T;
int ans[100],in[100],post[100],n;
void Build(int l1,int r1,int l2,int r2,Tree &T)
{
int i;
T=new node();
for(i=l1;i<=r1;i++)//求根结点
if(in[i]==post[r2])
break;
T->data=post[r2];
if(i==l1)
T->left=NULL;
else Build(l1,i-1,l2,l2+i-l1-1,T->left);//递归求左子树
if(i==r1)
T->right=NULL;
else Build(i+1,r1,r2-(r1-i),r2-1,T->right);//递归求右子树
}
int j;
void levelorder(Tree  T)
{
queue<Tree>q;
q.push(T);
while(!q.empty())
{
Tree root=q.front();
q.pop();
ans[j++]=root->data;
if(root->left!=NULL)
q.push(root->left);
if(root->right!=NULL)
q.push(root->right);
}
}
int main()
{
/*#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif*/
int i;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&post[i]);
for(i=1;i<=n;i++)
scanf("%d",&in[i]);
Build(1,n,1,n,T);
j=0;
levelorder(T);
for(i=0;i<j-1;i++)
printf("%d ",ans[i]);
printf("%d\n",ans[i]);
return 0;
}