PAT 1143 Lowest Common Ancestor[难][BST性质]

1143 Lowest Common Ancestor（30 分）

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 Ais 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

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.

题目大意：给出一棵二叉搜索树的前序遍历，并且给出两个节点，查询这两个节点的最近的共同祖先，如果其中一个是另一个的父节点，那么按格式输出，如果查不到该节点，那么根据相应的格式进行输出。即最小公共祖先。

//既然关键字的范围是int，那么就不能使用哈西father数组的形式来查找了。

//本来想用map,但是又考虑到会有重复的数，所以就不能用了。

PAT 1143 Lowest Common Ancestor[难][BST性质]-LMLPHP

View Code

//实在是不太会，就写了这么点，就是不知道怎么去给这些node标记父节点。

//看到柳神说这是水题，我的内心接受不了了。。

代码来自：https://www.liuchuo.net/archives/4616

#include <iostream>
#include <vector>
#include <map>
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;//表示这个节点出现了  } for (int i = 0; i < m; i++) { scanf("%d %d", &u, &v); for(int j = 0; j < n; j++) { a = pre[j];//其实每一个节点都是根节点。 if ((a >= u && a <= v) || (a >= v && a <= u)) break; //如果a在两者之间或者就是当前节点其中一个，  } if (mp[u] == false && mp[v] == false)//false就是都没有出现，也就是0。 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); } return 0; }

1.有一个规律，输入是按照前根遍历来输入的，那么每一个数的前一个数，就是当前数的根节点啊！哪里用建树呢？！

2.利用了搜索二叉树的性质，真是厉害，学习了。

3.判断a是在u和v之间，还是恰好是u和v.

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