PAT 甲级 1020 Tree Traversals (二叉树遍历）
PAT 甲级 1020 Tree Traversals (二叉树遍历）
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 7Sample Output:
4 1 6 3 5 7 2
根据后续遍历和中序遍历，求二叉树
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>
using namespace std;
typedef struct Tree
{
int data;
Tree *lchild;
Tree *rchild;
}a[40];
int post[40];
int in[40];
int n;
int ans[40];
void dfs(int l1,int r1,int l2,int r2,Tree* &root)
{
root=new Tree();
int i;
for( i=l1;i<=r1;i++)
if(in[i]==post[r2])
break;
root->data=post[r2];
if(i==l1)
root->lchild=NULL;
else
dfs(l1,i-1,l2,l2+i-l1-1,root->lchild);
if(i==r1)
root->rchild=NULL;
else
dfs(i+1,r1,r2-(r1-i),r2-1,root->rchild);
}
int cnt;
void bfs(Tree *tree)
{
queue<Tree*> q;
q.push(tree);
while(!q.empty())
{
Tree *root=q.front();
q.pop();
ans[cnt++]=root->data;
if(root->lchild!=NULL)
q.push(root->lchild);
if(root->rchild!=NULL)
q.push(root->rchild);
}
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&post[i]);
for(int i=1;i<=n;i++)
scanf("%d",&in[i]);
Tree *tree;
dfs(1,n,1,n,tree);
cnt=0;
bfs(tree);
for(int i=0;i<cnt;i++)
{
if(i==cnt-1)
printf("%d\n",ans[i]);
else
printf("%d ",ans[i]);
}
}
return 0;
}