【二叉树】用数组给出二叉树层序遍历序列，建树以及遍历问题

传递悄悄话

层序遍历数组形式的下标如下

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#include <algorithm>
#include <cstring>
#include <iostream>using namespace std;const int N = 1010, M = N * 2;int n;
int h[N], e[M], ne[M], idx;
int v[N], dist[N];
bool st[N];void add(int a, int b)
{e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}void dfs(int u)
{st[u] = true;for (int i = h[u]; i != -1; i = ne[i]){int j = e[i];if (!st[j]) {dist[j] = dist[u] + v[j];dfs(j);}}
}int main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);memset(h, -1, sizeof h);int x;while (cin >> x) v[++n] = x;for (int i = 1; i <= n / 2; i++)//二叉树的最后一个分支(非叶)节点为n/2{if (v[i] == -1) continue;int a = i * 2, b = i * 2 + 1;//若下标在范围内且左右儿子存在if (a <= n && v[a] != -1) add(i, a), add(a, i);//无向边if (b <= n && v[b] != -1) add(i, b), add(b, i);}dfs(1);//1号点为根节点int res = 0;//dist数组存的是根节点1到其他所有点的距离for (int i = 1; i <= n; i++) res = max(res, dist[i]);cout << res << '\n';return 0;
}