1020 Tree Traversals
1020 Tree Traversals (25 分)
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
题意:给出一颗二叉树的后序遍历和中序遍历,求这颗二叉树的层序遍历
理解上图的过程,这道题就好做了。
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=50;
struct node
{
int data;
node *lchild;
node *rchild;
};
int pre[maxn],in[maxn],post[maxn];//先序,中序,后续
int n;//结点个数
//当前二叉树的后序序列区间为[postL,postR]
//中序序列区间为[inL,inR]
node *create(int postL, int postR, int inL, int inR)
{
if(postL>postR)
{
return NULL;
}
node *root=new node;
root->data=post[postR];
int k;
for(k=inL;k<=inR;k++)
{
if(in[k]==post[postR])
break;
}
int numLeft=k-inL;
root->lchild=create(postL,postL+numLeft-1,inL,k-1);
root->rchild=create(postL+numLeft,postR-1,k+1,inR);
return root;
}
int num=0;
void BFS(node * root)
{
queue<node*> q;//队列里存地址
q.push(root);//根结点地址入队
while(!q.empty())
{
node * now=q.front();//取出队首元素
q.pop();
printf("%d",now->data);//访问队首元素
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(int i=0;i<n;i++)
{
scanf("%d", &post[i]);
}
for(int i=0;i<n;i++)
{
scanf("%d", &in[i]);
}
node * root=create(0,n-1,0,n-1);
BFS(root);
return 0;
}