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：

1.   给出m次查询与长度为n的前序序列。找到最近的公共祖先

 

思路：

      不用建树，用 前序+中序转层序的办法，每一次都等读到一个根，如果a 和b就位于当前根的左边和右边。那么就是这个根了，如果都落在左子树就朝左子树找，如果都落在右子树就朝右子树那边找。特别的如果a 或b与当前根是一致的，要特判。

参考代码：

 

 

 

#include<vector>
#include<algorithm>
#include<map>
#include<cstdio>
using namespace std;
vector<int> pre, in;
map<int, int> mp; 
void lca(int root, int st, int ed, int a, int b){
	if(st > ed) return;
	int now = mp[pre[root]], ra = mp[a], rb = mp[b];
	if((ra < now && rb > now) || (ra > now && rb < now))	printf("LCA of %d and %d is %d.\n", a, b, in[now]);
	else if(ra < now && rb < now)	lca(root + 1, st, now - 1, a, b);
	else if(ra > now && rb > now) lca(root + now - st + 1, now + 1, ed, a, b);
	else if(ra == now || rb == now){
		if(ra == now) swap(a, b);
		printf("%d is an ancestor of %d.\n", b, a);
	}
}
int n, m, a, b;
int main(){
	scanf("%d%d", &m, &n);
	pre.resize(n + 1); in.resize(n + 1);
	for(int i = 1; i <= n; ++i) scanf("%d", &in[i]), mp[in[i]] = i;
	for(int i = 1; i <= n; ++i)	scanf("%d", &pre[i]);
	for(int i = 0; i < m; ++i){
		scanf("%d%d", &a, &b);
		if(!mp[a] && !mp[b])	printf("ERROR: %d and %d are not found.\n", a, b);
		else if(!mp[a] || !mp[b]) printf("ERROR: %d is not found.\n", !mp[a]? a : b);
		else lca(1, 1, n, a, b);
	}
	return 0;
}

代码参考：https://blog.csdn.net/liuchuo/article/details/82560863