PAT-A1151 LCA in a Binary Tree 题目内容及题解

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.

题目大意

树中两个节点的最低公共祖先(LCA)是同时拥有该两个节点作为后代的最深节点。题目给定一棵二叉树及其中的任意两个节点，要求找到它们的LCA并按照格式要求输出。

解题思路

1. 初始化并读入树的中序遍历，用map容器建立数字与中序遍历位置信息的对应关系；
2. 读入树的前序遍历；
3. 对所有请求的节点对检查其是否在map内，否则用题目所给的格式输出未找到；
4. 如均能找到，则保存两节点在中序遍历中的位置信息；
5. 前序位置零开始判断直到前序位置末尾，如果出现当前位置的元素在中序遍历序列中的位置介于两节点的位置之间或者找到某一元素的情况时表明找到公共祖先，则按照格式输出未找到节点相关信息，并继续下一次查询；
6. 全部查询完成后返回零值。

代码

#include<cstdio>
#include<map>
using namespace std;
#define maxn 10010

int preOrder[maxn];

map<int,int> mp;
int M,N;

void Init(){
    int i,a;
    scanf("%d%d",&M,&N);
    for(i=0;i<N;i++){
        scanf("%d",&a);
        mp[a]=i;
    }
    for(i=0;i<N;i++){
        scanf("%d",&preOrder[i]);
    }
}

void Check(){
    int i,a,b,flag1=0,flag2=0;
    int sa,sb,sc;
    scanf("%d%d",&a,&b);
    if(mp.find(a)==mp.end()){
        flag1++;
    }
    if(mp.find(b)==mp.end()){
        flag2++;
    }
    if(flag1||flag2){
        if(flag1&&flag2){
            printf("ERROR: %d and %d are not found.\n",a,b);
        }else if(flag1){
            printf("ERROR: %d is not found.\n",a);
        }else{
            printf("ERROR: %d is not found.\n",b);
        }
        return;
    }
    sa=mp[a];
    sb=mp[b];
    for(i=0;i<N;i++){
        sc=mp[preOrder[i]];
        if(((sa<sc)&&(sb>sc))||((sa>sc)&&(sb<sc))){
            printf("LCA of %d and %d is %d.\n",a,b,preOrder[i]);
            return;
        }else if(sa==sc){
            printf("%d is an ancestor of %d.\n",a,b);
            return;
        }else if(sb==sc){
            printf("%d is an ancestor of %d.\n",b,a);
            return;
        }
    }
}

int main(){
    Init();
    while(M--){
        Check();
    }
    return 0;
}

运行结果