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.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

The lef subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the lef and right subtrees must also be binary search trees.

Given any two nodes in a BST, 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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. 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 BST, 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

6 3 1 2 5 4 8 7

2 5

8 7

1 9

12 -3

0 8

99 99

Sample Output:

LCA of 2 and 5 is 3.

8 is an ancestor of 7.

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

已知二叉查找树的前序序列，求每个测试样例中两个节点最近公共祖先节点

解题思路

利用二叉查找树特点：根节点，大于左子树所有结点，小于右子树所有结点

前序序列依次遍历，即为从左到右从下到上每个根节点，判断测试样例两个节点与每个根节点大小关系

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

易错点

利用前序升序排序得到中序，前序+中序查找lca会有一个测试点超时（对比A1151）

Code

#include <iostream>

#include <vector>

#include <map>

using namespace std;

map<int,bool> mp;

int main(int argc,char * argv[]) {

	int m,n,u,v,a;

	scanf("%d %d",&m,&n);

	vector<int> pre(n);

	for(int i=0; i<n; i++) {

		scanf("%d",&pre[i]);

		mp[pre[i]]=true;

	}

	for(int i=0; i<m; i++) {

		scanf("%d %d",&u,&v);

		for(int i=0; i<n; i++) {

			a=pre[i];

			if(a>=u&&a<=v||a<=u&&a>=v)break;

		}

		if(mp[u]==false&&mp[v]==false)  //都没找到

			printf("ERROR: %d and %d are not found.\n",u,v);

		else if(mp[u]==false||mp[v]==false)

			printf("ERROR: %d is not found.\n",mp[u]==false?u:v);

		else if(a==v||a==u)

			printf("%d is an ancestor of %d.\n",a,a==v?u:v);

		else

			printf("LCA of %d and %d is %d.\n",u,v,a);

	}

	return 0;

}

巴特西

PAT Advanced 1143 Lowest Common Ancestor (30) [二叉查找树 LCA]

题目

题目分析

解题思路

易错点

Code

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