PAT甲级1151（由前序和中序确定LCA）-白红宇

PAT甲级1151（由前序和中序确定LCA）

阅读量：5010 次

发布时间：2019-06-12

本文共 2733 字，大约阅读时间需要 9 分钟。

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 ( $\leq 1,000), the number of pairs of nodes to be tested; and N (\leq10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.$

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 87 2 3 4 6 5 1 85 3 7 2 6 4 8 12 68 17 912 -30 899 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. 前序和中序的解释：https://blog.csdn.net/ailunlee/article/details/80755357 代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            #include
             
              #include
              
               const int maxn=1e5+5;typedef long long ll;using namespace std;map
               
                pos;int in[maxn],pre[maxn];void LCA(int l,int r,int preroot,int a,int b){ if(l>r) { return; } int Root=pos[pre[preroot]]; int ina=pos[a]; int inb=pos[b]; if(ina
                
                 
                  Root&&inb>Root) { LCA(Root+1,r,preroot+Root-l+1,a,b); } else if ((ina < Root && inb> Root) || (ina > Root && inb < Root)) { printf("LCA of %d and %d is %d.\n", a, b, in[Root]); return ; } else if (ina == Root) { printf("%d is an ancestor of %d.\n", a, b); return ; } else if (inb == Root) { printf("%d is an ancestor of %d.\n", b, a); return ; }}int main(){ int m,n; cin>>m>>n; for(int t=1;t<=n;t++) { scanf("%d",&in[t]); pos[in[t]]=t; } for(int t=1;t<=n;t++) { scanf("%d",&pre[t]); } int a,b; while(m--) { scanf("%d%d",&a,&b); if(pos[a]==0&&pos[b]==0) { printf("ERROR: %d and %d are not found.\n",a,b); } else if(pos[a]==0&&pos[b]!=0) { printf("ERROR: %d is not found.\n",a); } else if(pos[a]!=0&&pos[b]==0) { printf("ERROR: %d is not found.\n",b); } else { LCA(1,n,1,a,b); } } return 0;}

转载于:https://www.cnblogs.com/Staceyacm/p/11361905.html

你可能感兴趣的文章

MySQL的外键，修改表，基本数据类型，表级别操作，其他（条件，通配符，分页，排序，分组，联合，连表操作）...