1151 LCA in a Binary Tree（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.

Given any two nodes in a binary tree, 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 binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. 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 binary tree, 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
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.
有一道考研題也是求lca，突然想到一個不用建樹的方法，不過要額外小心細節，避免超時。依然用map記錄元素在中序遍歷數組中位置，下標不為0，如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
int pre[10001],in[10001],m,n;
map<int,int> mp;
void check(int a,int b) {
    int aa = mp[a],bb = mp[b];
    if(!bb && !aa)printf("ERROR: %d and %d are not found.\n",a,b);
    else if(!aa)printf("ERROR: %d is not found.\n",a);
    else if(!bb)printf("ERROR: %d is not found.\n",b);
    else {
        if(a == b) {
            printf("%d is an ancestor of %d.\n",a,b);
            return;
        } 
        int p1 = 1,p2 = n,i1 = 1,i2 = n;
        while(p1 <= p2) {
            int temp = pre[p1],mtemp = mp[temp];
            if(temp == a) {
                printf("%d is an ancestor of %d.\n",a,b);
                return;
            }
            else if(temp == b) {
                printf("%d is an ancestor of %d.\n",b,a);
                return;
            }
            if(mtemp > aa && mtemp > bb) i2 = mtemp - 1,p1 ++,p2 = p1 + i2 - i1;
            else if(mtemp < aa && mtemp < bb) i1 = mtemp + 1,p1 = p2 - i2 + i1;
            else {
                printf("LCA of %d and %d is %d.\n",a,b,temp);
                return;
            }
        }
    }
}
int main() {
    int a,b;
    scanf("%d%d",&m,&n);
    for(int i = 1;i <= n;i ++) {
        scanf("%d",&in[i]);
        mp[in[i]] = i;
    }
    for(int i = 1;i <= n;i ++) {
        scanf("%d",&pre[i]);
    }
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        check(a,b);
    }
    return 0;
}

 

給定一棵樹的中序和前序遍歷，要求查詢兩個點的最低祖先，需要記錄高度。

#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;///根據序號映射到結點
tree *createNode(int d,int h)///創建新結點並返回
{
    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)///前中序 還原樹
{
    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;
}
void check(int a,int b)///判斷兩點
{
    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;
        }///調整至高度相同
        if(t1 == t2) {///一個點是另一個點的祖先
            printf("%d is an ancestor of %d.\n",t1 -> Data,a == t1 -> Data ? b : a);
            return;
        }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",&z[i]);
    }
    for(int i = 0;i < n;i ++) {
        scanf("%d",&q[i]);
    }
    head = createTree(0,n - 1,0,n - 1,0);
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        check(a,b);
    }
}