Pat grade a 1151 LCA in a binary tree (30 points)

1151 LCA in a Binary Tree (30  branch )

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.

   The main idea of the topic ：

1.   give m Query times and length is n The preorder sequence of . Find the nearest common ancestor

 

Ideas ：

      You don't have to build a tree , use Before the order + The method of transforming middle sequence into sequence , Every time I read a root , If a and b On the left and right of the current root . So this is the root , If it all falls on the left sub tree, look for it , If they all fall on the right subtree, look towards the right subtree . Special if a or b Consistent with the current root , A special sentence is required .

Reference code ：

 

 

 

#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;
}

Code reference ：https://blog.csdn.net/liuchuo/article/details/82560863