1020. Tree Traversals (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
提交代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int n,i,j;
struct node{
int data;
node*lchild;
node*rchild;
};
int post[50],in[50];
node*create(int postL,int postR,int inL,int inR)
{
if(postL>postR)
{
return NULL;
}
node*root = new node;//存放根节点的地址
root->data = post[postR];//根节点的值
//遍历找到左右子树的个数
int k;
for(k=inL;k<=inR;k++)
{
if(in[k]==post[postR])
break;
}
int numberLeft = k-inL;//左子树的个数
//赋值给左子树
root->lchild=create(postL,postL+numberLeft-1,inL,k-1);
root->rchild=create(postL+numberLeft,postR-1,k+1,inR);
return root;//返回根节点地址
}
int num=0;
void BFS(node*root)
{
queue<node*>q;
q.push(root);
while(!q.empty())
{
node*now=q.front();
q.pop();
printf("%d",now->data);
num++;
if(num<n){
printf(" ");
}
if(now->lchild!=NULL)q.push(now->lchild);
if(now->rchild!=NULL)q.push(now->rchild);
}
}
int main()
{
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&post[i]);
}
for(j=0;j<n;j++)
{
scanf("%d",&in[j]);
}
//建树
node*root = create(0,n-1,0,n-1);
//层序遍历
BFS(root);
return 0;
}
