[PAT] 1143 Lowest Common Ancestor（30 分）

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 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

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. 

题意：寻找二分查找树中两个节点的最深公共祖先。

思路：
1.先通过先序遍历和中序遍历得到树。
2.判断两个结点是否在树中。
3.求得两个结点的父节点。
4.对两个父节点最深公共父节点。

柳婼 の blog
1.遍历这个前序遍历
2.如果这两个节点分别在当前节点的左右字数，或者当前节点等于这两个节点中的一个，则得到最深公共祖先。

题解：

 1 #include<cstdio>
 2 #include<vector>
 3 #include<map>
 4 using namespace std;
 5 int main() {
 6     int m, n;
 7     map<int, bool> mp;
 8     scanf("%d %d", &m, &n);
 9     vector<int> pre(n);
10     for (int i = 0; i<n; i++){
11         scanf("%d", &pre[i]);
12         mp[pre[i]] = true;
13     }
14     int u, v;
15     for (int i = 0; i < m; i++) {
16         scanf("%d %d", &u, &v);
17         int a;
18         for (int j = 0; j < pre.size(); j++) {
19             a = pre[j];
20             if ((a < u && a > v) || (a > u && a < v) || (a == u) || (a == v)) {
21                 break;
22             }
23         }
24         if (mp[u] == false && mp[v] == false) {
25             printf("ERROR: %d and %d are not found.\n", u, v);
26         }
27         else if(mp[u] == false || mp[v] == false) {
28             printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
29         }
30         else if (a == u || a == v) {
31             printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
32         }
33         else {
34             printf("LCA of %d and %d is %d.\n", u, v, a);
35         }
36     }
37     return 0;
38 }