PAT-甲级-1151 LCA in a Binary Tree (30 分)

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.

题目大意：给定中序和先序，找出m组查询中，a、b结点的公共祖先，并按照给定输出格式输出。

解题思路：

• 用vector分别保存中序in、先序pre；
• 用unordered_map<int,int> 保存中序元素对应的下标；
• 通过in和pre，用dfs 遍历树；
• LCA的判断条件是：
  • aIn：a在中序对应的下标；bIn：b在中序对应的下标；i：当前根结点u在中序对应的下标；
  • a、b分别在树的两侧：(aIn < i && bIn > i ) || (aIn > i && bIn < i )
  • a、b都在树的左侧，此时遍历左子树：(aIn < i && bIn < i )；dfs(u + 1 , l , i-1 , a , b)；
  • a、b都在树的右侧，此时遍历右子树：(aIn > i && bIn > i )；dfs(u - l + i + 1 , i+1 , r , a , b)；
  • aIn与i相等、或者b与i相等：相等的结点即为另一个不等的祖先结点。

#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i,j,k)    for(int i=j;i<k;i++)
vector<int> pre,in;
unordered_map<int,int> pos;
void dfs(int u,int l,int r,int a,int b){
    if(l>r)    return;
    int i = pos[pre[u]],aIn = pos[a],bIn = pos[b];
    if(aIn<i && bIn<i)
        dfs(u+1,l,i-1,a,b);
    else if((aIn>i && bIn<i) || (aIn<i && bIn>i))
        printf("LCA of %d and %d is %d.\n",a,b,in[i]);
    else if(aIn>i && bIn>i)
        dfs(u-l+i+1,i+1,r,a,b);
    else if(aIn==i)
        printf("%d is an ancestor of %d.\n",a,b);
    else if(bIn==i)
        printf("%d is an ancestor of %d.\n",b,a);
}
int main(){
    std::ios::sync_with_stdio(false);
    int m,n,a,b;
    cin>>m>>n;
    in.resize(n+1);
    pre.resize(n+1);
    rep(i,1,n+1){
        cin>>in[i];
        pos[in[i]] = i;
    }
    rep(i,1,n+1)
    cin>>pre[i];
    rep(i,0,m){
        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|| pos[b]==0)
            printf("ERROR: %d is not found.\n",pos[a]==0?a:b);
        else
            dfs(1,1,n,a,b);
    }
    return 0;
}