PAT--1151 LCA in a Binary Tree--LCA

題目鏈接：https://pintia.cn/problem-sets/994805342720868352/problems/1038430130011897856

題目大意：給定中序和先序，求給定兩個點的最近公共祖先。

分析：假設給出的是a和b，若a和b在當前根節點的左子樹上，那麼就在當前根節點的左子樹上找，同理，都在右子樹上，就在當前根節點的右子樹上找，如果一個左邊，一個右邊，那麼答案就是當前根節點。

14行：inroot - l是左子樹的節點數，所以先序數組肯定往右移動inroot - l，然後再加上當前根節點，所以再加1。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int n, m, in[N], pre[N];
map<int, int> pos;
void lca(int l, int r, int preroot, int a, int b) {
	if(l > r) return ;
	int inroot = pos[pre[preroot]], ain = pos[a], bin = pos[b];
	if(ain < inroot && bin < inroot) {
		lca(l, inroot - 1, preroot + 1, a, b);
	} else if((ain < inroot && bin > inroot) || (ain > inroot && bin < inroot)) {
		printf("LCA of %d and %d is %d.\n", a, b, in[inroot]);
	} else if(ain > inroot && bin > inroot) {
		lca(inroot + 1, r, preroot + 1 + inroot - l, a, b);
	} else if(ain == inroot) {
		printf("%d is an ancestor of %d.\n", a, b);
	} else if(bin == inroot) {
		printf("%d is an ancestor of %d.\n", b, a);
	}
}
int main() {
	scanf("%d %d", &m, &n);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &in[i]);
		pos[in[i]] = i;
	}
	for(int i = 1; i <= n; i++)
		scanf("%d", &pre[i]);
	int a, b;
	while(m--) {
		scanf("%d %d", &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 {
			lca(1, n, 1, a, b);
		}
	}
	return 0;
}