C++

$\color{gold}{— > 蓝桥杯辅导课题解}$

思路：

贪心 DFS

$节点的最大高度 = 子节点的数量 + 子节点所能形成的最大深度$

$code1:$

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;

int n;
int h[N], e[N], ne[N], idx;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

int dfs(int u) {
    int hmax = 0, cnt = 0; // cnt:子节点的数量 hmax:子节点所能形成的最大深度
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        hmax = max(hmax, dfs(j));
        cnt ++;
    }
    return hmax + cnt;
}

int main() {
    cin >> n;
    memset(h, -1, sizeof h);
    for (int i = 2; i <= n; i ++) {
        int p; cin >> p;
        add(p, i); // 在邻接表中加入父节点和当前点的边
    }
    cout << dfs(1);
    return 0;
}

$code2:$

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;

int n;
vector<int> g[N];

int dfs(int u) {
    int ans = 0, cnt = g[u].size();
    for (int i = 0; i < cnt; i ++)
        ans = max(ans, dfs(g[u][i]) + cnt);

    return ans;    
}

int main() {
    cin >> n;
    for (int i = 2; i <= n; i ++) {
        int x; cin >> x;
        g[x].push_back(i);
    }

    cout << dfs(1);
    return 0;
}