数据结构实验之二叉树四:(先序中序)还原二叉树
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
给定一棵二叉树的先序遍历序列和中序遍历序列,要求计算该二叉树的高度。
Input
输入数据有多组,每组数据第一行输入1个正整数N(1 <= N <= 50)为树中结点总数,随后2行先后给出先序和中序遍历序列,均是长度为N的不包含重复英文字母(区分大小写)的字符串。
Output
输出一个整数,即该二叉树的高度。
Sample Input
9 ABDFGHIEC FDHGIBEAC
Sample Output
5
Hint
Source
xam\
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char pre[55];
char mid[55];
struct node
{
char data;
struct node *left;
struct node *right;
};
struct node *creat(char *pre, char *mid, int n)
{
if(n <= 0) return NULL;
int k;
struct node *root;
root = (struct node *)malloc(sizeof(struct node));
root -> data = *pre;
for(k = 0; k < n; k++)
{
if(mid[k] == pre[0])
break;
}
root -> left = creat(pre + 1, mid, k);
root -> right = creat(pre + k + 1, mid + k + 1, n - k - 1);
return root;
};
int height(struct node *root) // 递归求高度
{
if(root)
{
int m = height(root -> left);
int n = height(root -> right);
return (m > n)?(m + 1):(n + 1);
}
else
return 0;
}
int main()
{
int h;
int n;
while(~scanf("%d", &n))
{
scanf("%s", pre);
scanf("%s", mid);
struct node *root;
root = creat(pre, mid, n);
h = height(root);
printf("%d\n", h);
}
return 0;
}