L2-011 玩转二叉树 (25 分)
摘要
Title: L2-011 玩转二叉树 (25 分)
Tag: 二叉树
Memory Limit: 64 MB
Time Limit: 1000 ms
Powered by:NEFU AB-IN
L2-011 玩转二叉树 (25 分)
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题意
给定一棵二叉树的中序遍历和前序遍历,请你先将树做个镜面反转,再输出反转后的层序遍历的序列。所谓镜面反转,是指将所有非叶结点的左右孩子对换。这里假设键值都是互不相等的正整数。
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思路
求镜面翻转的层序遍历,就是在最后bfs时右叶节点先入队即可
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代码
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72/*
* @Author: NEFU AB-IN
* @Date: 2022-04-20 22:34:10
* @FilePath: \ACM\GPLT\L2-011.CPP
* @LastEditTime: 2022-04-20 22:44:30
*/
using namespace std;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;
const int N = 60;
deque<int> q;
int preorder[N], inorder[N], n;
unordered_map<int, int> pos, lt, rt;
vector<int> v;
int build(int il, int ir, int pl, int pr)
{
int root = preorder[pl];
int k = pos[root];
if (il < k)
lt[root] = build(il, k - 1, pl + 1, pl + 1 + k - 1 - il);
if (ir > k)
rt[root] = build(k + 1, ir, pl + 1 + k - 1 - il + 1, pr);
return root;
}
void bfs(int root)
{
q.push_front(root);
while (q.size())
{
auto root = q.back();
q.pop_back();
v.push_back(root);
if (rt[root])
q.push_front(rt[root]);
if (lt[root])
q.push_front(lt[root]);
}
}
signed main()
{
IOS;
cin >> n;
for (int i = 0; i < n; ++i)
{
cin >> inorder[i];
pos[inorder[i]] = i;
}
for (int i = 0; i < n; ++i)
{
cin >> preorder[i];
}
int root = build(0, n - 1, 0, n - 1);
bfs(root);
for (int i = 0; i < SZ(v); ++i)
{
cout << v[i] << " "[i == SZ(v) - 1];
}
return 0;
}