1127 ZigZagging on a Tree (30 分)

蛇形输出层序遍历，这个只要记录每一层的遍历序列，然后输出之前该翻转的翻转一下就可以了，然后看了看自己一年之前提交的，思路真的是一模一样，就是代码风格变了，

#include <bits/stdc++.h>

#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pa;

struct node {
    int val, lchild, rchild, level;
} tree[35];

int tot, maxlevel;
int post[35], in[35];
vi v[35];
map<int, int> mp;

int create(int postl, int postr, int inl, int inr) {
    if (postl > postr) return -1;
    int root = tot++; tree[root].val = post[postr];
    int k = mp[post[postr]];
    tree[root].lchild = create(postl, postl + k - inl - 1, inl, k - 1);
    tree[root].rchild = create(postl + k - inl, postr - 1, k + 1, inr);
    return root;
}

void layerorder(int root) {
    queue<int> q;
    q.push(root);
    tree[root].level = 0;
    while (!q.empty()) {
        int now = q.front(); q.pop();
        int l = tree[now].lchild, r = tree[now].rchild;
        int level = tree[now].level;
        v[level].pb(tree[now].val);
        if (level > maxlevel) maxlevel = level;
        if (l != -1) { q.push(l); tree[l].level = level + 1; }
        if (r != -1) { q.push(r); tree[r].level = level + 1; }
    }
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) { cin >> in[i]; mp[in[i]] = i; }
    for (int i = 0; i < n; i++) cin >> post[i];
    int root = create(0, n - 1, 0, n - 1);
    layerorder(0); int ok = 0;
    for (int i = 0; i <= maxlevel; i++) {
        if (i && i % 2 == 0) reverse(all(v[i]));
        for (int j = 0; j < v[i].size(); j++) cout << (ok++ ? " " : "") << v[i][j];
    }
    return 0;
}

标签：now,level,int,tree,30,Tree,postl,ZigZagging,root
来源： https://blog.csdn.net/weixin_46709784/article/details/118634368