PTA: 浪漫倒影 [二叉树重建] [深度优先遍历]

题目

输入样例：

8
6 8 7 4 5 1 3 2
8 5 4 7 6 3 2 1

输出样例：

R: 1 2 3 4 5
L: 1 6 7 8 5

思路

重建二叉树，然后深度优先遍历，如果是看右视图，就优先遍历右子树；如果是左视图，就优先遍历左子树。遍历过程中记录每个深度是否有节点已经被看到，打印出第一个被看到的节点。

代码

#include <cstdio>
#include <cstring>

using namespace std;

struct node {
    
    int n;
    node* l, *r;
    node() {
    
        l = r = NULL;
    }
};

int last[25], mid[25];

node* build(int mb, int me, int lb, int le) {
    
    if(mb > me || lb > le) return NULL;
    node* t = new node();
    t->n = last[le];
    int m;
    for(int i = mb; i <= me; ++i) {
    
        if(mid[i] == last[le]) {
    
            m = i;
            break;
        }
    }
    int ls = m-mb;
    t->l = build(mb, m-1, lb, lb+ls-1);
    t->r = build(m+1, me, lb+ls, le-1);
    return t;
}
bool vis[20];
void right(node* tree, int depth) {
    
    if(tree) {
    
        if(!vis[depth]) {
    
            printf(" %d", tree->n);
            vis[depth] = true;
        }
        right(tree->r, depth+1);
        right(tree->l, depth+1);
    }
}
void left(node* tree, int depth) {
    
    if(tree) {
    
        if(!vis[depth]) {
    
            printf(" %d", tree->n);
            vis[depth] = true;
        }
        left(tree->l, depth+1);
        left(tree->r, depth+1);
    }
}

int main(void)
{
    
    freopen("in.txt", "r", stdin);
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) {
    
        scanf("%d", &mid[i]);
    }
    for(int i = 0; i < n; ++i) {
    
        scanf("%d", &last[i]);
    }
    node* tree = build(0, n-1, 0, n-1);
    memset(vis, 0, sizeof(vis));
    printf("R:");
    right(tree, 0);
    printf("\n");
    memset(vis, 0, sizeof(vis));
    printf("L:");
    left(tree, 0);
    return 0;
}