1151. LCA in a Binary Tree

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.

分析

这个题目就是将前中序遍历构建二叉树与lca算法结合起来进行考察。这两个问题拆开来看其实都不算困难，但是结合起来以后有一定技巧。最直接的思路就是首先将二叉树构建起来，然后再dfs进行搜索即可。但是其实不需要这样麻烦，只需要在“构建”的同时进行判断即可。因此可以抽象出一个框架，对于前中序遍历确定二叉树的问题，只要保持其构建的框架不变，然后在其中添加必要的操作即可，不一定要把树构建出来。

代码

//
// Created by masterCai on 2020/5/7.
//

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

using namespace std;

// 因c++中不能直接找到元素对应下标，因此用map映射一下方便后续查找元素对应位置
map<int, int> pos; 
vector<int> in, pre; //存储中序，前序序列

void lca(int inl, int inr, int preRoot, int a, int b){
    //inl：中序left inr中序right preroot：当前的root a，b：待查询的数据
    if(inl > inr){ // 中序全部遍历完 退出
        return;
    }
    //从pos中找到当前的root的位置 查找a，b在中序中的位置
    int inRoot = pos[pre[preRoot]], aIn = pos[a], bIn = pos[b];
    if(aIn < inRoot && bIn < inRoot) { // a，b都在左边 搜索左子树即可
        lca(inl, inRoot - 1, preRoot + 1, a, b);
    } else if((aIn < inRoot && bIn>inRoot) || (aIn > inRoot && bIn < inRoot)){// 若a，b在两边 则找到答案了
        printf("LCA of %d and %d is %d.\n", a, b, in[inRoot]);
    } else if(aIn > inRoot && bIn>inRoot){ // 都在右边 搜索右子树
        lca(inRoot+1, inr, preRoot+1+(inRoot-inl), a, b);
    } else if(aIn == inRoot){ // a或b是根，则找到答案
        printf("%d is an ancestor of %d.\n", a, b);
    } else if(bIn == inRoot){
        printf("%d is an ancestor of %d.\n", b, a);
    }
}


int main(){
    int m, n;
    cin >> m >> n;
    in.resize(n+1), pre.resize(n+1);
    // 这里小技巧：在全局声明没有大小的vector，然后在得到大小后resize
    //（直接全局声明vector<int> a(n)）这个意思。

    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]);
    }
    int a, b;
    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &a, &b);
        if (pos[a] == 0 && pos[b] == 0){ //如果都没找到
            printf("ERROR: %d and %d are not found.\n", a, b);
        } else if(pos[a] == 0 || pos[b] == 0){ //没找到其中一个
            printf("ERROR: %d is not found.\n", pos[a] == 0?a:b);
        } else{
            lca(1, n, 1, a, b); //搜索即可
        }
        return 0;
    }
    return 0;
}