1143 Lowest Common Ancestor（附测试点2，3段错误分析）

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 (

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.

思路：

1、利用二叉搜索树的性质和前序遍历建树，即通过中序遍历和前序遍历建树（二叉搜索树的中序遍历就是从小到大的排序）

2、然后开始前序遍历（假设需要找u,v两点的祖先）可以分为下面几个情况
①u和v在root的两侧，也就是一个大于root一个小于root那么root就是他们的共同祖先。
②若u，v都小于root那么进行遍历root的左子树，反之遍历其右子树。

③同时要判断u和v不存在的情况

3、测试点2，3段错误

因为key的值的范围为int，因此需要使用map或者unordered_map来表示key是否存在

代码：

#include<stdio.h>
#include<map>
using namespace std;
int preorder[10005];
map<int, bool> f;
int M, N;
struct Node{
    int x;
    Node* lchild;
    Node* rchild;
};
Node* build(int left, int right){
    if(left > right){
        return NULL;
    }
    if(left == right){
        Node* node = new Node;
        node -> x = preorder[left];
        node -> lchild = NULL;
        node -> rchild = NULL;
        return node;
    }
    int k = right + 1;
    Node* node = new Node;
    node -> x = preorder[left];
    for(int i = left + 1; i <= right; i++){ // 不要忘记考虑不存在preorder[i] > preorder[left]的情况，此时k未被赋值
        if(preorder[i] > preorder[left]){
            k = i;
            break;
        }
    }
    node -> lchild = build(left + 1, k - 1);
    node -> rchild = build(k, right);
    return node;
}
 void find(Node* node, int x, int y){
     if(x < node -> x && y < node -> x){
         find(node -> lchild, x, y);
     }else if(x > node -> x && y > node -> x){
         find(node -> rchild, x, y);
     }else if(node -> x == x){
         printf("%d is an ancestor of %d.\n", x, y);
         return;
     }else if(node -> x == y){
         printf("%d is an ancestor of %d.\n", y, x);
         return;
     }
     else{
         printf("LCA of %d and %d is %d.\n", x, y, node -> x);
         return;
     }
 }
int main(){
    scanf("%d%d", &M, &N);
    for(int i = 1; i <= N; i++){
        scanf("%d", &preorder[i]);
        f[preorder[i]] = true;
    }
    Node* root = build(1, N);
    for(int i = 0; i < M; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        if(!f[x] && !f[y]){
            printf("ERROR: %d and %d are not found.\n", x, y);
            continue;
        }else if(!f[x] && f[y]){
            printf("ERROR: %d is not found.\n", x);
            continue;
        }else if(f[x] && !f[y]){
            printf("ERROR: %d is not found.\n", y);
            continue;
        }
        find(root, x, y);
    }
    return 0;
}