PAT Advanced 1151 LCA in a Binary Tree (30) [树的遍历，LCA算法]

题目

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.

题目分析

已知二叉树中序和前序序列，求每个测试用例两个节点的最近公共祖先节点

解题思路

二叉树中序+前序唯一确定一棵二叉树，利用（前序第一个节点root在中序中将中序分为左子树和右子树）查找LCA

1. u，v在root两边，则root为u，v最近公共祖先节点
2. u，v都在root左边，则最近公共祖先在root左子树，递归查找
3. u，v都在root右边，则最近公共祖先在root右子树，递归查找
4. u==root，则u是v的最近公共祖先节点
5. v==root，则v是u的最近公共祖先节点

Code

#include <iostream>
#include <map>
#include <vector>
using namespace std;
vector<int> pre,in;
map<int,int> pos;
void lca(int inL,int inR,int preL,int u,int v) {
	if(inL>inR)return;
	int rin=pos[pre[preL]], uin=pos[u], vin=pos[v];
	if((uin<rin&&vin>rin)||(vin<rin&&uin>rin)) {
		printf("LCA of %d and %d is %d.\n",u,v,in[rin]);
	} else if(uin<rin&&vin<rin) { //u,v在左子树
		lca(inL, rin-1, preL+1,u,v);
	} else if(uin>rin&&vin>rin) { //u,v在右子树
		lca(rin+1, inR, preL+(rin-inL)+1,u,v);
	} else if(uin==rin) {  //u是v lca
		printf("%d is an ancestor of %d.\n",u,v);
	} else if(vin==rin) {  //v是u lca
		printf("%d is an ancestor of %d.\n",v,u);
	}
}
int main(int argc,char * argv[]) {
	int m,n,u,v;
	scanf("%d %d",&m,&n);
	pre.resize(n+1);
	in.resize(n+1);
	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]);
	}
	for(int i=0; i<m; i++) {
		scanf("%d %d",&u,&v);
		if(pos[u]==0&&pos[v]==0) { //都没找到
			printf("ERROR: %d and %d are not found.\n",u,v);
		} else if(pos[u]==0||pos[v]==0) {
			printf("ERROR: %d is not found.\n",pos[u]==0?u:v);
		} else {
			lca(1,n,1,u,v);
		}
	}
	return 0;
}