DFS树的遍历

1.完全二叉树的遍历，$数组index从1开始，左Child = parent * 2, 右Child = parent * 2 + 1$
2.到达叶结点，打印搜索路径

C++ 代码

#include <iostream>
using namespace std;

const int MAX_N = 1010;
int heap[MAX_N];
int path[MAX_N];
int n;
bool isMinHeap, isMaxHeap;
void DFS(int v, int step) {
    path[step] = heap[v];
    if (v * 2 > n) {
        for (int i = 0; i <= step; i++) {
            if (i) { 
                if (path[i] > path[i - 1]) isMinHeap = true;
                else if (path[i] < path[i - 1]) isMaxHeap = true;
                printf(" "); 
            }
            printf("%d", path[i]);
        }
        cout << endl;
        return;
    }
    if (v * 2 + 1 <= n) { 
        DFS(v * 2 + 1, step + 1); 
    }
    if (v * 2 <= n) { 
        DFS(v * 2, step + 1); 
    }
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> heap[i];
    }
    DFS(1, 0);
    if (isMaxHeap && isMinHeap) puts("Not Heap");
    else if (isMaxHeap) puts("Max Heap");
    else puts("Min Heap");
    return 0;
}