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Search in a Binary Search Tree
To search a key in a binary search tree, we start from the root and move all the way down, choosing branches according to the comparison results of the keys. The searching path corresponds to a sequence of keys. For example, following {1, 4, 2, 3} we can find 3 from a binary search tree with 1 as its root. But {2, 4, 1, 3} is not such a path since 1 is in the right subtree of the root 2, which breaks the rule for a binary search tree. Now given a sequence of keys, you are supposed to tell whether or not it indeed correspnds to a searching path in a binary search tree.
首先 理解题意 给出一个序列 每一个数都是前一个数的左右儿子之一 保证这棵树是搜索树
这道题其实是考察搜索二叉树的建立 我用了个取巧的方法 并没有建二叉树 凡是满足题意的序列 有一个规律
若第二个数比第一个数大 那后面的所有数都比第一个数大 若第三个数比第二个数小 那第三个数以后的数都比第二个数小
以{1,4,2,3}为例子 大家可以验证一下
至于为什么有这个规律 大家想一下搜索二叉树的定义就知道了
下面给出AC代码
1 #include "stdio.h" 2 main() 3 { 4 int m,n; 5 scanf("%d%d",&m,&n); 6 int i,j,k,flag; 7 int a[100]; 8 for(i=0;i<m;i++) 9 { 10 flag=0; 11 for(j=0;j<n;j++) 12 { 13 scanf("%d",&a[j]); 14 } 15 for(k=0;k<n-1;k++) 16 { 17 if(a[k+1]>a[k]) 18 { 19 for(j=k+1;j<n;j++) 20 { 21 if(a[j]<=a[k]) 22 { 23 flag=-1;break; 24 } 25 } 26 } 27 else 28 { 29 for(j=k+1;j<n;j++) 30 { 31 if(a[j]>=a[k]) 32 { 33 flag=-1; 34 break; 35 } 36 } 37 } 38 } 39 if(flag==-1) 40 printf("NO\n"); 41 else 42 printf("YES\n"); 43 } 44 45 }
PAT Search in a Binary Search Tree
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原文地址：http://www.cnblogs.com/threezj/p/4525014.html
