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Tree Traversals

题目链接：1020 Tree Traversals (25 point(s))

题干大意

给定一棵二叉树的节点数、后续遍历序列、中序遍历序列，要求输出其层序遍历序列

思路

中序遍历可以确定左右子树，后续遍历可以确定子树的根，先构造出二叉树，然后做层序遍历即可

AC代码

C++
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#include <iostream>
#include <queue>
#include <vector>
using namespace std;

struct node {
    int          data;
    struct node* left;
    struct node* right;
};
int         n;
vector<int> post; // NOLINT
vector<int> in;   // NOLINT
vector<int> layer;

node* construct(int in_l, int in_r, int post_l, int post_r)
{
    if (in_r - in_l < 1) {
        return nullptr;
    }
    int   key  = post_r - 1;
    node* root = new node;
    root->data = post[key];
    int i      = in_l;
    for (; i != in_r; ++i) {
        if (in[i] == post[key])
            break;
    }
    root->left  = construct(in_l, i, post_l, post_r - in_r + i);
    root->right = construct(i + 1, in_r, post_r - in_r + i, post_r - 1);
    return root;
}

void traverse(node*& tree)
{
    if (!tree)
        return;
    queue<node*> q;
    q.push(tree);
    while (!q.empty()) {
        node* front = q.front();
        q.pop();
        layer.emplace_back(front->data);
        if (front->left)
            q.push(front->left);
        if (front->right)
            q.push(front->right);
    }
}

int main()
{
    cin >> n;
    post.resize(n);
    in.resize(n);
    for (auto& it : post)
        cin >> it;
    for (auto& it : in)
        cin >> it;
    node* tree = construct(0, n, 0, n);
    traverse(tree);
    for (auto& it : layer) {
        cout << it;
        if (&it != &layer.back()) {
            cout << " ";
        }
    }
    return 0;
}