题目描述

给定一棵二叉树的后序遍历和中序遍历，请你输出其层序遍历的序列。这里假设键值都是互不相等的正整数。

思路

根据后序遍历和中序遍历结果重构二叉树后进行层序遍历即可

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 50;
int postorder[N], inorder[N];
map<int, int> mp;//记录中序遍历中每个节点的位置
vector<int> ans;
int n;
struct Node
{
    int val;
    Node *left;
    Node *right;
    Node(int x)
    {
        val = x;
        left = nullptr;
        right = nullptr;
    }
};

Node *rec(int pl, int pr, int il, int ir)
{
    if(pl > pr)
        return nullptr;
    Node *root = new Node(postorder[pr]);//后序遍历的最后一个点是根节点
    int k = mp[postorder[pr]];//找到中序遍历根节点位置
    auto left = rec(pl, pl + k - il - 1, il, k - 1);
    auto right = rec(pl + k - il ,pr - 1, k + 1, ir);
    root->left = left;
    root->right = right;
    return root;
}

Node *bulid()
{
    for (int i = 1; i <= n; i++)
        mp[inorder[i]] = i;
    Node *root = rec(1, n, 1, n);//重建树
    return root;
}

void bfs(Node *root)
{
    queue<Node *> q;
    q.push(root);
    while(q.size())
    {
        Node *now = q.front();
        q.pop();
        if(now == nullptr)
            continue;
        ans.push_back(now->val);
        auto left = now->left;
        auto right = now->right;
        q.push(left);
        q.push(right);
    }
    return;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> postorder[i];
    for (int i = 1; i <= n; i++)
        cin >> inorder[i];
    Node *root = bulid();
    bfs(root);
    for (int i = 0; i < ans.size(); i++)
    {
        cout << ans[i];
        if(i != ans.size() - 1)
            cout << ' ';
    }
    cout << endl;
    return 0;
}