1151 LCA in a Binary Tree-PAT甲级

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.

最近公共祖先（LCA）利用倍增解法来求解，在建树的过程中更新结点的父节点以及深度值。

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=10005;
struct tree{
	int data;
	tree *lchild=NULL;
	tree *rchild=NULL;
};
struct node{
	//结点类
	int data,father=-1,level;//权值，父节点在pre中的下标、深度 
	node(){}
	node(int d,int f,int l){
		data=d;
		father=f;
		level=l;
	}
};
map<int,int>mp;//建立值在前序序列中的下标 
node pre[N];
int m,n,in[N];
tree *create(int prel,int prer,int inl,int inr,int father,int level){
	if(prel>prer){
		return NULL;
	}
	tree *root=new tree;
	root->data=pre[prel].data;
	int k;
	for(k=inl;k<=inr;k++){
		if(in[k]==pre[prel].data)
			break;
	}
	int numleft=k-inl;
	pre[prel]=node(pre[prel].data,father,level);
	root->lchild=create(prel+1,prel+numleft,inl,k-1,prel,level+1);//递归处理左子树 
	root->rchild=create(prel+numleft+1,prer,k+1,inr,prel,level+1);//递归处理右子树 
    return root;
}
int main(){
	scanf("%d%d",&m,&n);
	for(int i=0;i<n;i++){
		scanf("%d",&in[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%d",&pre[i].data);
		mp[pre[i].data]=i;
	}
	tree *root=create(0,n-1,0,n-1,-1,0);
	for(int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		int idx,idy;
		if(mp.find(x)==mp.end()) idx=-1;
		else idx=mp[x];
		if(mp.find(y)==mp.end()) idy=-1;
		else idy=mp[y];
		if(idx==-1&&idy==-1){
			printf("ERROR: %d and %d are not found.\n",x,y);
		}else if(idx==-1){
			printf("ERROR: %d is not found.\n",x);
		}else if(idy==-1){
			printf("ERROR: %d is not found.\n",y);
		}else{
			int flag=1;//flag=1表示x的深度比b的深度大
			if(pre[idx].level<pre[idy].level){
				swap(idx,idy);//令idx指向更深的深度
				flag=0; 	
			} 
			while(pre[idx].level>pre[idy].level){
				idx=pre[idx].father;
			}
			if(idx==idy){
				printf("%d is an ancestor of %d.\n",pre[idy].data,flag?x:y);
			}else{
				while(idx!=idy){
					//二者同时向上调整；
					idx=pre[idx].father;
					idy=pre[idy].father; 
				}
				printf("LCA of %d and %d is %d.\n",x,y,pre[idx].data);
			}
		}
		
	}
	return 0;
}