1151 LCA in a Binary Tree（30 point(s)）

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.

//AC code
/*
  1:根據中序，先序建樹 
  2: 用 findnode( )檢視待查詢節點是否存在
  3:若待查詢節點是存在,用 LCA( )查詢最近公共祖先 
 其中LCA有三種： 
    u,v都在左子樹或右子樹,或者u,v分別在左子樹和右子樹，
 因此可採取如下步驟：
    (1)::判斷當前遍歷的節點是否為空，為空返回null，
    (2)::節點資料域是否等於u，是否等於p，是的話返回當前節點。
    (3)::之後判斷left 和right是否為空， 若都不為空 ，則當前root為祖先指標，
        若其中一個為空，則返回另一邊，為答案。
  4:格式輸出 
*/
 
 
#include<cstdio>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
const int maxn=10010;
struct node{
    int data;
    node *lchild,*rchild;
};

int pre[maxn],in[maxn];
int n,k,u,v;

node *creat(int prel,int prer,int inl,int inr){
    if(prel>prer){
        return NULL;
    }
    node *root=new node;
    root->data=pre[prel]; 
    int k;
    for( k=inl;k<=inr;k++){
        if(in[k]==pre[prel])
        break;
    }  
    int leftnum=k-inl; 
    root->lchild =creat(prel+1,prel+leftnum,inl,k-1); //下標要寫對哦
    root->rchild =creat(prel+leftnum+1,prer,k+1,inr);
    return root;
}

bool findnode(int u){ //查詢節點 
    for(int i=0;i<n;i++){
        if(u==pre[i])
         return true;
    }
    return false;
}

node *LCA(node *root, int u, int v) { //查詢兩節點最近祖先 
        if (root == NULL) return NULL;
        if (root->data== u || root->data== v) return root;
        node *left =LCA(root->lchild,u,v);
        node *right = LCA(root->rchild,u,v);
        if (left && right ) return root; //u,v分別位於左右子樹的情況
        return left == NULL ? right : left;
}

int main(){
    scanf("%d %d",&k,&n);
    for(int i=0;i<n;i++)
        scanf("%d",&in[i]);
    for(int i=0;i<n;i++)
        scanf("%d",&pre[i]);
    node *root=creat(0,n-1,0,n-1); // 尷尬，剛開始中序，先序順序輸反了，調了半天bug 
    for(int i=0;i<k;i++){
        scanf("%d %d",&u,&v);
        if(findnode(u)==false && findnode(v)==false)
           printf("ERROR: %d and %d are not found.\n", u, v);
        else if (findnode(u)==false || findnode(v)==false)
            printf("ERROR: %d is not found.\n", findnode(u)==false ? u : v);
        else {
            node *temp=LCA(root,u,v);
            if(temp->data==u || temp->data==v){
               printf("%d is an ancestor of %d.\n",temp->data==u? u:v,temp->data==u? v:u);
            }else{
                printf("LCA of %d and %d is %d.\n",u,v,temp->data);
            }
        }
    }
    return 0;
}