最低公共祖先（LCA）

对于一颗给定的二叉树，对任意两个节点，求它们的最低的公共祖先。

下面就是这样一道题，给定一棵树的前序和中序遍历（我们知道这颗树就是确定的了），对于任两个节点，求它们的最低公共主祖先。

#include<iostream>
#include<cstdio>
#include<map>
using namespace std;

int* pre,*in;
map<int,int> pos;

void LCA(int preRoot,int inL,int inR,int u,int v){
    if(inL>inR) return;
    //inRoot,u,v:根节点,u,v在中序遍历的位置
    int inRoot=pos[pre[preRoot]],uInIdx=pos[u],vInIdx=pos[v];
    if(uInIdx<inRoot&&vInIdx<inRoot){//u,v在中序遍历中的位置小于根 
        LCA(preRoot+1,inL,inRoot-1,u,v);
    }else if((uInIdx<inRoot&&vInIdx>inRoot)
            ||(uInIdx>inRoot&&vInIdx<inRoot)){
            printf("LCA of %d and %d is %d.\n",u,v,in[inRoot]);
    }else if(uInIdx>inRoot&&vInIdx>inRoot){
        LCA(preRoot+1+(inRoot-inL),inRoot+1,inR,u,v);
    }else{
        if(uInIdx==inRoot){
            printf("%d is an ancestor of %d.\n",u,v);
        }else if(vInIdx==inRoot){
            printf("%d is an ancestor of %d.\n",v,u);
        }
    }
}
int main(){
    int m,n;
    scanf("%d%d",&m,&n);
    in=new int[n+1];
    pre=new int[n+1];
    for(int i=1;i<=n;i++){
        scanf("%d",&in[i]);
        pos[in[i]]=i;
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&pre[i]);
    }
    while(m--){
        int a,b;
        scanf("%d%d",&a,&b);
        if(pos[a]==0&&pos[b]==0){
            printf("ERROR: %d and %d are not found.\n",a,b);
        }else if(pos[a]==0){
            printf("ERROR: %d is not found.\n",a);
        }else if(pos[b]==0){
            printf("ERROR: %d is not found.\n",b);
        }else{
            LCA(1,1,n,a,b);
        }
    }
    return 0;
}

上面是对于一颗普通的二叉树，如果这颗二叉树又是一个二叉搜索树，那么还有如下的方法可用，而且是更好的方式，用上面的方法就会超时。

题目：1143 Lowest Common Ancestor （30 分）

#include <iostream>
#include <algorithm>
#include <unordered_set>
#define MAXSIZE 10001
using namespace std;

struct Node{
    int val;
    Node *lchild, *rchild;
    Node(int v){
        val = v;
        lchild = rchild = NULL;
    }
};

int pre[MAXSIZE] = {0}, in[MAXSIZE] = {0};

Node * inPreCreateTree(int inL, int inR, int preL, int preR){
    if( preL > preR ){
        return NULL;
    }

    int val = pre[preL];
    Node * root = new Node(val);

    int mid;
    for( mid = inL; mid <= inR; mid++ ){
        if( in[mid] == val ){
            break;
        }
    }
    int numLeft = mid - inL;
    root->lchild = inPreCreateTree(inL, mid-1, preL+1, preL+numLeft);
    root->rchild = inPreCreateTree(mid+1, inR, preL+numLeft+1, preR);
    return root;
}

void LCA(Node * root, int lVal, int rVal, bool positionChanged){
    if( root != NULL ){
        if( root->val > lVal && root->val < rVal ){
            if( positionChanged == false ){
                cout<<"LCA of "<<lVal<<" and "<<rVal<<" is "<<root->val<<".\n";
            }else{
                cout<<"LCA of "<<rVal<<" and "<<lVal<<" is "<<root->val<<".\n";
            }
        }else if( root->val == lVal ){
            cout<<lVal<<" is an ancestor of "<<rVal<<".\n";
        }else if( root->val == rVal ){
            cout<<rVal<<" is an ancestor of "<<lVal<<".\n";
        }else if( root->val > rVal ){
            LCA(root->lchild, lVal, rVal, positionChanged);
        }else{
            LCA(root->rchild, lVal, rVal, positionChanged);
        }
    }
}

int main(){
//  ios::sync_with_stdio(false);
//  cin.tie(0);

    int m, n;
    cin>>m>>n;

    int val;
    unordered_set<int> s;
    for( int i = 0; i < n; i++ ){
        cin>>val;
        in[i] = pre[i] = val;
        s.insert(val);
    }

    sort(in, in+n);

    Node * root = inPreCreateTree(0, n-1, 0, n-1);

    int u, v;
    for( int i = 0; i < m; i++ ){
        cin>>u>>v;
        if( s.find(u) == s.end() && s.find(v) == s.end() ){
            cout<<"ERROR: "<<u<<" and "<<v<<" are not found.\n";
        }else if( s.find(u) == s.end() ){
            cout<<"ERROR: "<<u<<" is not found.\n";
        }else if( s.find(v) == s.end() ){
            cout<<"ERROR: "<<v<<" is not found.\n";
        }else{
            bool positionChanged = false;
            if( u >= v ){
                swap(u, v);
                positionChanged = true;
            }
            LCA(root, u, v, positionChanged);
        }
    }

    return 0;
}

方法2：

该方法对于这题会有一个测试点超时

#include<iostream>
#include<algorithm>
#include<map>
using namespace std;

const int maxn=10010;
int pre[maxn],in[maxn];
map<int,int> pos;

void LCA(int preRoot,int inL,int inR,int u,int v){
    if(inL>inR) return;
    int uInIdx=pos[u],vInIdx=pos[v];
    int inRoot=pos[pre[preRoot]];
    if(uInIdx<inRoot&&vInIdx<inRoot){
        LCA(preRoot+1,inL,inRoot-1,u,v);
    }else if(uInIdx<inRoot&&vInIdx>inRoot||(uInIdx>inRoot&&vInIdx<inRoot)){
        printf("LCA of %d and %d is %d.\n",u,v,pre[preRoot]);
    }else if(uInIdx>inRoot&&vInIdx>inRoot){
        LCA(preRoot+(inRoot-inL)+1,inRoot+1,inR,u,v);
    }else{
        if(uInIdx=inRoot){
            printf("%d is an ancestor of %d.\n",u,v);
        }else if(vInIdx==inRoot){
            printf("%d is an ancestor of %d.\n",v,u);
        }
    }
}
int main(){
    int m,n;
    cin>>m>>n;
    for(int i=1;i<=n;i++){
        cin>>pre[i];
        in[i]=pre[i];   
    }
    sort(in+1,in+n+1);
    for(int i=1;i<=n;i++){
        pos[in[i]]=i;
    }
    while(m--){
        int a,b;
        cin>>a>>b;
        if(pos[a]==0&&pos[b]==0){
            printf("ERROR: %d and %d are not found.\n",a,b);
        }else if(pos[a]==0){
            printf("ERROR: %d is not found.\n",a);
        }else if(pos[b]==0){
            printf("ERROR: %d is not found.\n",b);
        }else{
            LCA(1,1,n,a,b);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/sqmax/p/10410821.html