PAT 甲级 1151 LCA in a Binary Tree（30 分） C++版

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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.

这个题目把LCA算法和前序+中序转树的算法结合起来考了。 关于LCA算法可以练练LeetCode上面两个关于这个算法的题目：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/ 和 https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/ 在PAT的题库里也有LCA和二叉搜索树结合的题目: https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 将一棵树抽象为三个结点，分别为根结点和左子树结点以及右子树结点，那么a和b的最近公共祖先有三种状态，即a和b都在左子树结点或者都在右子树结点、a和b分别在左右子树结点、a或者b是当前根结点，对于第一种状态，最近公共祖先就可以在左子树或者右子树进行查找了，这样递归进去又是一颗新的树；对于第二种状态，当前根结点就是a和b的公共祖先；对于第三种状态，a(b)是b(a)的祖先。 代码的主体就是一个前序+中序转后序的算法，只要在每一层判断在中序序列中，当前层的根结点和a与b两个结点的状态（LCA）即可。

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#include <cstdio>
#include <map>

using namespace std;

map<int, int> pos;

int *in, *pre;

void lca(int inl, int inr, int pre_root, int a, int b) {
    if (inl > inr) return;
    int in_root = pos[pre[pre_root]], a_in_index = pos[a], b_in_index = pos[b];
    if (a_in_index < in_root && b_in_index < in_root) {
        lca(inl, in_root - 1, pre_root + 1, a, b);
    } else if ((a_in_index < in_root && b_in_index > in_root) ||
               (a_in_index > in_root && b_in_index < in_root)) {
        printf("LCA of %d and %d is %d.\n", a, b, in[in_root]);
    } else if (a_in_index > in_root && b_in_index > in_root) {
        lca(in_root + 1, inr, pre_root + 1 + (in_root - inl), a, b);
    } else {
        if (a_in_index == in_root) {
            printf("%d is an ancestor of %d.\n", a, b);
        } else if (b_in_index == in_root) {
            printf("%d is an ancestor of %d.\n", b, a);
        }
    }
}

int main() {
    int m = 0, n = 0, a = 0, b = 0;
    scanf("%d %d", &m, &n);
    in = new int[n + 1];
    pre = new int[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", &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) {
            printf("ERROR: %d is not found.\n", a);
        } else if (pos[b] == 0) {
            printf("ERROR: %d is not found.\n", b);
        } else {
            lca(1, n, 1, a, b);
        }
    }
    return 0;
}