PAT A1151 LCA in a Binary Tree (30point(s))

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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.

题意

给出二叉树的中序和前序遍历序列,判断任意两个结点是否在二叉树上,如果都在二叉树上,则求离它们最近的祖先。

解答

本题不需要我们建树,但需要用到建树的思想。由前序和中序序列可以唯一确定一棵树,对于某个前序序列,该序列的第一个键一定是根结点,根据该根结点的值可以由中序序列求出左右子树对应的中序和前序遍历序列,因此可以利用该方法遍历所有结点。如果给定的两个结点在当前结点的两边,则该结点一定是我们要求的祖先,如果都在左边,则从左子树寻找,否则从右子树寻找。解答本题时要控制好时间,减少不必要的运算,寻找前序值在中序序列的位置时,可以用map获取,判断结点是否在树上时也可以用map来判断。

#include<iostream>
#include<map>
#include<vector>

using namespace std;

int M, N, u, v;

vector<int> pre, in;
map<int, bool> exists;
map<int, int> inIndex;

void lca(int prel, int inl, int inr)
{
	int root = pre[prel];
	int uIn = inIndex[u], vIn = inIndex[v], rootIn = inIndex[pre[prel]];
	if ((uIn < rootIn && vIn > rootIn) || 
		(vIn < rootIn && uIn > rootIn))
		printf("LCA of %d and %d is %d.\n", u, v, root);
	else if (uIn == rootIn || vIn == rootIn)
		printf("%d is an ancestor of %d.\n", root, uIn == rootIn ? v : u);
	else if (uIn < rootIn && vIn < rootIn)
		lca(prel + 1, inl, rootIn - 1);// 到左树找
	else
		lca(prel + rootIn - inl + 1, rootIn + 1, inr);// 到右树找
}

int main()
{
	scanf("%d %d", &M, &N);
	pre.resize(N);
	in.resize(N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &in[i]);
		inIndex[in[i]] = i;
	}
	for (int i = 0; i < N; i++) {
		scanf("%d", &pre[i]);
		exists[pre[i]] = true;
	}
	
	for (int i = 0; i < M; i++) {
		scanf("%d %d", &u, &v);
		if (!exists[u] && !exists[v]) printf("ERROR: %d and %d are not found.\n", u, v);
		else if (!exists[u] && exists[v]) printf("ERROR: %d is not found.\n", u);
		else if (exists[u] && !exists[v]) printf("ERROR: %d is not found.\n", v);
		else lca(0, 0, N - 1);
	}
	return 0;
}