求二叉树的先序遍历
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
已知一棵二叉树的中序遍历和后序遍历,求二叉树的先序遍历
Input
输入数据有多组,第一行是一个整数t (t<1000),代表有t组测试数据。每组包括两个长度小于50 的字符串,第一个字符串表示二叉树的中序遍历序列,第二个字符串表示二叉树的后序遍历序列。
Output
输出二叉树的先序遍历序列
Sample Input
2 dbgeafc dgebfca lnixu linux
Sample Output
abdegcf xnliu
Hint
Source
GYX
AC代码
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
char mid[51] , hou[51];
typedef struct node
{
char data;
struct node *left , *right;
} tree;
tree *creat(int n , char *mid , char *hou)
{
if (n == 0)
return NULL;
int i;
tree *root;
root = (tree *)malloc(sizeof(tree));
root ->data = hou[n-1];
for (i = 0 ; i < n ; i++)
{
if (hou[n-1] == mid[i])
{
break;
}
}
root->left = creat(i , mid , hou);
root ->right = creat(n - i - 1 , mid + i + 1 , hou + i);
return root;
}
void front(tree *root)
{
if (root != NULL)
{
printf("%c" , root->data);
front(root ->left);
front(root ->right);
}
}
int main()
{
int t;
tree *root;
scanf("%d" , &t);
while(t--)
{
int n;
scanf("%s" , mid);
scanf("%s" , hou);
n = strlen(mid);
root = (tree *)malloc(sizeof(tree));
root = creat(n , mid , hou);
front(root);
printf("\n");
}
return 0;
}
/***************************************************
User name: jk170717
Result: Accepted
Take time: 0ms
Take Memory: 172KB
Submit time: 2018-08-08 19:42:27
****************************************************/
超内存代码
1 :
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
char mid[51] , hou[51];
typedef struct node
{
char data;
struct node *left , *right;
} tree;
tree *creat(int n , char *mid , char *hou)
{
if (n == 0)
return NULL;
int i;
tree *root;
root = (tree *)malloc(sizeof(tree));
root ->data = hou[n-1];
for (i = 0 ; i < n ; i++)
{
if (hou[n-1] == mid[i])
{
break;
}
}
root->left = creat(n , mid , hou);
root ->right = creat(n - i - 1 , mid + i + 1 , hou + i);
return root;
}
void front(tree *root)
{
if (root != NULL)
{
printf("%c" , root->data);
front(root ->left);
front(root ->right);
}
}
int main()
{
int t;
tree *root;
scanf("%d" , &t);
while(t--)
{
int n;
scanf("%s" , mid);
scanf("%s" , hou);
n = strlen(mid);
root = (tree *)malloc(sizeof(tree));
root = creat(n , mid , hou);
front(root);
printf("\n");
}
return 0;
}
/***************************************************
User name: jk170717
Result: Memory Limit Exceeded
Take time: 0ms
Take Memory: 65536KB
Submit time: 2018-08-08 19:41:26
****************************************************/
2:
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
char mid[51] , hou[51];
typedef struct node
{
char data;
struct node *left , *right;
} tree;
tree *creat(int n , char *mid , char *hou)
{
if (n == 0)
return NULL;
int i;
tree *root;
root = (tree *)malloc(sizeof(tree));
root ->data = hou[n-1];
for (i = 0 ; i < n ; i++)
{
if (hou[n-1] == mid[i])
{
break;
}
}
root->left = creat(n , mid , hou);
root ->right = creat(n - i - 1 , mid + i + 1 , hou + i);
return root;
}
void front(tree *root)
{
if (root)
{
printf("%c" , root->data);
front(root ->left);
front(root ->right);
}
}
int main()
{
int t;
int n;
tree *root;
scanf("%d" , &t);
while(t--)
{
scanf("%s" , mid);
scanf("%s" , hou);
n = strlen(mid);
root = (tree *)malloc(sizeof(tree));
root = creat(n , mid , hou);
front(root);
printf("\n");
}
return 0;
}
/***************************************************
User name: jk170717
Result: Memory Limit Exceeded
Take time: 0ms
Take Memory: 65536KB
Submit time: 2018-08-08 19:38:13
****************************************************/