PAT（甲级）1143. Lowest Common Ancestor（30）
PAT（甲级）1143. Lowest Common Ancestor（30）
PAT 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: 1.The left subtree of a node contains only nodes with keys less than the node’s key. 2.The right subtree of a node contains only nodes with keys greater than or equal to the node’s key. 3.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. 输入格式： 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 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.
输出格式：
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..
输入样例：
6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99输出样例：
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.题目分析：题目给出的是BST树，可以利用BST的中序遍历是顺序序列这一性质。由前序序列排序后获得中序序列，使用中序序列和前序序列重构二叉树即可。
AC代码：
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
const int maxn = 10010;
map<int, bool> mp;
int in[maxn];
int pre[maxn];
int post[maxn];
int preTemp[maxn];
struct node{
int data;
node* lchild;
node* rchild;
};
node* build(int preL, int preR, int inL, int inR){
if(preL>preR)return NULL;
node* root = new node;
root->data = pre[preL];
int k;
for(k=inL; k<=inR; ++k){
if(in[k] == pre[preL])break;
}
int leftnum = k-inL;
root->lchild = build(preL+1, preL+leftnum, inL, k-1);
root->rchild = build(preL+leftnum+1, preR, k+1, inR);
return root;
}
void postorder(node* root){
if(root){
postorder(root->lchild);
postorder(root->rchild);
cout<<root->data<<" ";
}
}
node* lca(node* root, int u, int v){
if(root==NULL)return NULL;
if(root->data==u || root->data==v)return root;
node* left = lca(root->lchild, u, v);
node* right = lca(root->rchild, u, v);
if(left && right)return root;
return left==NULL?right:left;
}
int main(){
int m, n, u, v;
scanf("%d%d",&m,&n);
for(int i=0; i<n; ++i){
scanf("%d", &pre[i]);
mp[pre[i]] = true;
preTemp[i] = pre[i];
}
sort(preTemp,preTemp+n);
for(int i=0; i<n; ++i){in[i]=preTemp[i];}
node* root = build(0, n-1, 0, n-1);
for(int i=0; i<m; ++i){
scanf("%d%d",&u,&v);
if(mp[u]==false && mp[v]==false)
printf("ERROR: %d and %d are not found.\n", u, v);
else if(mp[u]==false || mp[v]==false){
if(mp[u]==false)
printf("ERROR: %d is not found.\n",u);
else
printf("ERROR: %d is not found.\n",v);
}
else{
node* tmp = lca(root, u, v);
if(tmp->data==u || tmp->data==v){
if(tmp->data == u)
printf("%d is an ancestor of %d.\n", u, v);
else
printf("%d is an ancestor of %d.\n", v, u);
}
else{
printf("LCA of %d and %d is %d.\n", u, v, tmp->data);
}
}
}
return 0;
}腾讯云开发者
