PAT 甲级 1151  LCA in a Binary Tree

1151 LCA in a Binary Tree （30 point(s)）

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

Given any two nodes in a binary tree, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line ​​LCA of U and V is A.​​​ if the LCA is found and ​​A​​​ is the key. But if ​​A​​​ is one of U and V, print ​​X is an ancestor of Y.​​​ where ​​X​​​ is ​​A​​​ and ​​Y​​​ is the other node. If U or V is not found in the binary tree, print in a line ​​ERROR: U is not found.​​​ or ​​ERROR: V is not found.​​​ or ​​ERROR: U and V are not found.​​.

Sample Input:

6 8
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

经验总结：

emmmm  比前一个LCA​​PAT 甲级 1143 Lowest Common Ancestor​​难了一点 ，主要是，前一个是二叉搜索树，这一个就是普通的二叉树，无法根据结点的值之间的大小关系直接寻找LCA，而且这一题给的是中序序列以及前序序列，需要在输入的时候记录每一个结点在中序序列中的位置，然后基于中序和前序建树的函数进行修改，利用根结点在中序中的位置以及查询的两个数在中序中位置的关系，判断属于哪种情况，这一题也是参考了柳神的代码。。。。学习到了0.0

AC代码

#include <cstdio>
#include <vector>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
const int maxn=10010;
int m,n,pre[maxn],in[maxn],a,b;
map<int,int> ppos,ipos;

int main()
{
  scanf("%d%d",&m,&n);
  for(int i=1;i<=n;++i)
  {
    scanf("%d",&in[i]);
    ipos[in[i]]=i;
  }
  for(int i=1;i<=n;++i)
  {
    scanf("%d",&pre[i]);
    ppos[pre[i]]=i;
  }
  for(int i=0;i<m;++i)
  {
    scanf("%d%d",&a,&b);
    if(ipos[a]==false&&ipos[b]==false)
      printf("ERROR: %d and %d are not found.\n",a,b);
    else if(ipos[a]==false||ipos[b]==false)
      printf("ERROR: %d is not found.\n",ipos[a]==false?a:b);
    else
    {
      if(abs(ipos[a]-ipos[b])==1)
        printf("%d is an ancestor of %d.\n",ppos[a]<ppos[b]?a:b,ppos[a]<ppos[b]?b:a);
      else
      {
        int L=ipos[a]>ipos[b]?ipos[b]:ipos[a];
        int R=ipos[a]>ipos[b]?ipos[a]:ipos[b];
        int f=false;
        for(int j=L+1;j<R;++j)
          if(ppos[in[j]]<ppos[a]&&ppos[in[j]]<ppos[a])
          {
            printf("LCA of %d and %d is %d.\n",a,b,in[j]);
            break;
          }
      }
    }
  }
  return 0;
}