PAT 1151 LCA in a Binary Tree[难][二叉树]

1151 LCA in a Binary Tree （30 分）

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 Ais 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.

 题目大意：给出一个二叉树的中根遍历和前根遍历，然后给出查询，查询这两个节点的最低公共子节点。

 //当时看到题目之后，就想到了之前有一道类似的，不过那个二叉树是二叉搜索树。

//使用前根和中根可以确定后根遍历结果，不用重建二叉树，然后根据前根和后根，假设查询节点为u和v

//比如前根和后跟遍历中，既在u/v节点的前面又在其后面，那么肯定就是u/v的祖父节点，然后比较二者的祖父节点，查看最低相同的。

//当时的思路是这样，不过只得了22分，有一个测试点是超时的，不知道怎么改。

代码来自：https://www.liuchuo.net/archives/6496

#include <iostream>
#include <vector>
#include<cstdio>
#include <map>
using namespace std;
map<int, int> pos;
vector<int> in, pre;
void lca(int inl, int inr, int preRoot, int a, int b) {
    if (inl > inr) return;
    int inRoot = pos[pre[preRoot]], aIn = pos[a], bIn = pos[b];
    if (aIn < inRoot && bIn < inRoot)//在左子树。
        lca(inl, inRoot-1, preRoot+1, a, b);
    else if ((aIn < inRoot && bIn > inRoot) || (aIn > inRoot && bIn < inRoot))
        printf("LCA of %d and %d is %d.
", a, b, in[inRoot]);
        //一个在当前左，一个在当前右。
    else if (aIn > inRoot && bIn > inRoot)
        lca(inRoot+1, inr, preRoot+1+(inRoot-inl), a, b);
    else if (aIn == inRoot)
            printf("%d is an ancestor of %d.
", a, b);
    else if (bIn == inRoot)
            printf("%d is an ancestor of %d.
", b, a);
}
int main() {
    int m, n, a, b;
    scanf("%d %d", &m, &n);
    in.resize(n + 1), pre.resize(n + 1);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &in[i]);
        pos[in[i]] = i;//它原来是个map啊。并不是什么遍历序列，只是用来判断是否出现过。
    }
    for(int i=1;i<=n;i++){
        printf("%d ",pos[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.
", a, b);
        else if (pos[a] == 0 || pos[b] == 0)
            printf("ERROR: %d is not found.
", pos[a] == 0 ? a : b);
        else
            lca(1, n, 1, a, b);
    }
    return 0;
}

//这个pos映射很奇怪啊，不知道是什么意思呢，是第几个出现的吗？像是二叉树从左到右，从下到上的这个顺序，不太理解。总之好厉害，需要学习。