进阶实验8-2.3 二叉搜索树的最近公共祖先 (30分)

给定一棵二叉搜索树的先序遍历序列，要求你找出任意两结点的最近公共祖先结点（简称 LCA）。

输入格式：

输入的第一行给出两个正整数：待查询的结点对数 M（≤ 1 000）和二叉搜索树中结点个数 N（≤ 10 000）。随后一行给出 N 个不同的整数，为二叉搜索树的先序遍历序列。最后 M 行，每行给出一对整数键值 U 和 V。所有键值都在整型int范围内。

输出格式：

对每一对给定的 U 和 V，如果找到 A 是它们的最近公共祖先结点的键值，则在一行中输出 LCA of U and V is A.。但如果 U 和 V 中的一个结点是另一个结点的祖先，则在一行中输出 X is an ancestor of Y.，其中 X 是那个祖先结点的键值，Y 是另一个键值。如果 二叉搜索树中找不到以 U 或 V 为键值的结点，则输出 ERROR: U is not found. 或者 ERROR: V is not found.，或者 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.

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
struct tree
{
    int Data,Height;
    tree *Last,*Left,*Right;
}*head;
int q[10001],z[10001],m,n;
map<int,tree *> mp;
tree *createNode(int d,int h)
{
    tree *p = new tree();
    p -> Data = d;
    mp[d] = p;
    p -> Height = h;
    p -> Last = p -> Left = p -> Right = NULL;
    return p;
}
tree *createTree(int ql,int qr,int zl,int zr,int h)
{
    tree *p = createNode(q[ql],h);
    for(int i = zl;i <= zr;i ++)
    {
        if(z[i] == q[ql])
        {
            if(i > zl)p -> Left = createTree(ql + 1,ql + i - zl,zl,i - 1,h + 1),p -> Left -> Last = p;
            if(i < zr)p -> Right = createTree(ql + i - zl + 1,qr,i + 1,zr,h + 1),p -> Right -> Last = p;
            break;
        }
    }
    return p;
}
tree *createTre(int l,int r,int h)
{
    tree *p = createNode(q[l],h);
    for(int i = l + 1;i <= r + 1;i ++)
    {
        if(i == r + 1 || q[i] >= q[l])
        {
            if(i > l + 1)p -> Left = createTre(l + 1,i - 1,h + 1),p -> Left -> Last = p;
            if(r >= i)p -> Right = createTre(i,r,h + 1),p -> Right -> Last = p;
            return p;
        }
    }
}
void check(int a,int b)
{
    if(mp[a] == NULL && mp[b] == NULL)printf("ERROR: %d and %d are not found.\n",a,b);
    else if(mp[a] == NULL)printf("ERROR: %d is not found.\n",a);
    else if(mp[b] == NULL)printf("ERROR: %d is not found.\n",b);
    else
    {
        tree *t1 = mp[a],*t2 = mp[b];
        while(t1 -> Height != t2 -> Height)
        {
            if(t1 -> Height > t2 -> Height)t1 = t1 -> Last;
            else t2 = t2 -> Last;
        }
        if(t1 == t2)
        {
            printf("%d is an ancestor of %d.\n",t1 -> Data,a == t1 -> Data ? b : a);
            return;
        }
        t1 = t1 -> Last;
        t2 = t2 -> Last;
        while(t1 != t2)
        {
            t1 = t1 -> Last;
            t2 = t2 -> Last;
        }
        printf("LCA of %d and %d is %d.\n",a,b,t1 -> Data);
    }
}
int main()
{
    int a,b;
    scanf("%d%d",&m,&n);
    for(int i = 0;i < n;i ++)
    {
        scanf("%d",&q[i]);
    }
    head = createTre(0,n - 1,0);
    for(int i = 0;i < m;i ++)
    {
        scanf("%d%d",&a,&b);
        check(a,b);
    }
}