单词积累
respectively 分别地 各自地
descendants 后代 晚辈 子节点
题目
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 (≤ 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.
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 99
结尾无空行
Sample 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.
结尾无空行
思路
求二叉排序树的最低公共父节点，常规思路是从根节点开始遍历，遍历到第一个能把两个数从中间分开的节点，就找到了结果。但是该条思路有两个测试样例（3和4）超时了。不过作为一般思路，仍有借鉴意义。
通过研究题解，发现有更巧妙的做法，无需建树。题目给定了前序序列和待查询数a和b，遍历前序序列，找到第一个大小位于a和b间的数，就是结果。该做法全篇无树，但又巧用树的性质，解决了树的问题。
为什么是第一个呢？因为前序序列，遇到的首先是根，这个根会把后续的树分成左右子树，a和b也会被分开，不会再出现共同的父节点。
代码（AC版本）
#include<bits/stdc++.h>
using namespace std;
map<int, bool> mp;
int main() {
int m, n, u, v, a;
scanf("%d %d", &m, &n);
vector<int> pre(n);
for (int i = 0; i < n; i++) {
scanf("%d", &pre[i]);
mp[pre[i]] = true;
}
while (m--) {
scanf("%d %d", &u, &v);
for (int j = 0; j < n; j++) {
a = pre[j];
if ((a >= v && a <= u) || (a >= u && a <= v)) break;
}
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)
printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
else if (a == u || a == v)
printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
else
printf("LCA of %d and %d is %d.\n", u, v, a);
}
}
代码2（常规超时版）
#include <bits/stdc++.h>
using namespace std;
int N, M;
const int maxn = 10003;
int num[maxn];
map<int,int> vis;
vector<int> ans;
struct node{
int data;
node* lchild;
node* rchild;
node(int d): data(d), lchild(NULL), rchild(NULL) {
}
};
void insert(node* &root, int v) {
if (root == NULL) {
root = new node(v);
return;
}
if (v < root->data) {
insert(root->lchild, v);
} else if (v >= root->data) {
insert(root->rchild, v);
}
}
void search(node* root, int a, int b) {
if (a == root->data) {
printf("%d is an ancestor of %d.\n", a, b);
return;
}else if (b == root->data) {
printf("%d is an ancestor of %d.\n", b, a);
return;
}else if (a < root->data && b < root->data) {
search(root->lchild, a, b);
} else if (a > root->data && b > root->data) {
search(root->rchild, a, b);
} else {
printf("LCA of %d and %d is %d.\n", a, b, root->data);
}
}
int main() {
cin>>N>>M;
node* root = NULL;
for (int i = 0; i < M; i++) {
cin>>num[i];
vis[num[i]] = 1;
insert(root, num[i]);
}
while (N--) {
int n1, n2;
cin>>n1>>n2;
if (vis[n1] == 1 && vis[n2] == 1) {
search(root, n1, n2);
} else if (vis[n1] == 1) {
printf("ERROR: %d is not found.\n", n2);
} else if (vis[n2] == 1) {
printf("ERROR: %d is not found.\n", n1);
} else {
printf("ERROR: %d and %d are not found.\n", n1, n2);
}
}
}
