1143. Lowest Common Ancestor (30)
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.
A binary search tree (BST) is recursively defined as a binary tree which has the following properties:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
- Both the left and right subtrees must also be binary search trees.
Given any two nodes in a BST, 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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. 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 BST, 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 6 3 1 2 5 4 8 7 2 5 8 7 1 9 12 -3 0 8 99 99Sample Output:
LCA of 2 and 5 is 3. 8 is an ancestor of 7. ERROR: 9 is not found. ERROR: 12 and -3 are not found. ERROR: 0 is not found. ERROR: 99 and 99 are not found.
The meaning of the question is to find the nearest common ancestor of U and V points in the BST tree as required and output them as required
We can traverse the entire table directly. Since the pre-order sequence of BST is the insertion order of elements, which is the order of depth-first search of elements, we can directly scan the entire table, which is equivalent to a deep search of the entire tree. If the current point meets the output conditions, the output will be stopped.
But then I have been wrong and I don't know why I finally referred to the code on the Internet and found that even if I found the LCA of two points, I would output U first and then V in the input order.
Hey. . . I have been looking for this mistake for a long time. PAT is not friendly to input and output at all. . .
#include<bits/stdc++.h>
using namespace std;
int a[10010];
map<int,bool>Map;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++){
scanf("%d",&a[i]);
Map[a[i]]=1;
}
while(n--){
int s,e;
scanf("%d%d",&s,&e);
bool fs=0,fe=0;
fs = Map[s],fe = Map[e];
if(fs==0&&fe==0)
printf("ERROR: %d and %d are not found.\n",s,e);
else if(fs!=0&&fe==0)
printf("ERROR: %d is not found.\n",e);
else if(fs==0&&fe!=0)
printf("ERROR: %d is not found.\n",s);
else{
int ss = s,ee=e;
s = min(ss,ee),e = max(ee,ss);
for(int now = 1;now<=m;now++){
if((s<a[now]&&e>a[now])){
printf("LCA of %d and %d is %d.\n",ss,ee,a[now]);//This is the output here If the output of the modified element of s,e will lead to two tests click error
break;
}
if(s==a[now]){
printf("%d is an ancestor of %d.\n",s,e);
break;
}
if(e==a[now]){
printf("%d is an ancestor of %d.\n",e,s);
break;
}
}
}
}
return 0;
}