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由二叉树的后序和中序求层次遍历

2016-08-14 10:50 232 查看
PAT 1020

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:给出后序和中序,求出层次遍历
#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
const int maxn=50;
struct node                     //节点
{
int date;
node* ld;
node* rd;
} ;
int pre[maxn],in[maxn],post[maxn];
int n;         //节点个数
node* create(int postl,int postr,int inl,int inr)    //由后序和中序构建一颗树
{
if(postl>postr)
return NULL;
node* root=new node;              //建立一个新的节点,存放当前二叉树的根节点
root->date =post[postr];          //新节点的数据域为当前根节点的值
int k;
for(k=inl;k<=inr;k++)
{
if(in[k]==post[postr])
break;
}
int numleft=k-inl;              //左子树的节点个数
root->ld =create(postl,postl+numleft-1,inl,k-1);
root->rd =create(postl+numleft,postr-1,k+1,inr);
return root;
}
int num=0;                          //已经输出的节点个数,便于空格输出
void bfs(node* root)
{
queue<node*> q;
q.push(root);
while(!q.empty()  )
{
node* now=q.front() ;
q.pop() ;
printf("%d",now->date );
num++;
if(num<n) printf(" ");
if(now->ld !=NULL) q.push(now->ld );
if(now->rd !=NULL) q.push(now->rd );

}
}
int main()
{
scanf("%d",&n);
for(int i=0;i<
4000
;n;i++)   //后序输入
scanf("%d",&post[i]);
for(int i=0;i<n;i++)  //中序输入
scanf("%d",&in[i]);
node* root=create(0,n-1,0,n-1);
bfs(root);           //层序输出
return 0;
}
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