题目

题目大意

给出一棵树的后序遍历和中序遍历，要求输出该树的层序遍历。

思路

非常典型的树的构建与遍历问题。后序遍历和中序遍历可以得出一个树的结构，用递归锁定根节点，然后再遍历左右子树，我之前发过类似题目的博客，这里就不再详细赘述。层序遍历就是使用队列了。

代码

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

int n;
vector<int> hou, mid, level;
struct node{
    int data;
    node * lchild, *rchild;
};

void buildtree(node * &root, int hl, int hr, int ml, int mr){
    if (hl > hr || ml > mr){
        return;
    }
    int pos;
    for (int i = ml; i <= mr; i++){
        if (hou[hr] == mid[i]){
            pos = i;
            break;
        }
    }
    root = new node();
    root->data = hou[hr];
    root->lchild = root->rchild = nullptr;
    buildtree(root->lchild, hl, hl + pos - ml - 1, ml, pos - 1);
    buildtree(root->rchild, hl + pos - ml, hr - 1, pos + 1, mr);
}

void findlevel(node * root){
    queue<node *> q;
    q.push(root);
    while (!q.empty()){
        node * now = q.front();
        level.push_back(now->data);
        q.pop();
        if (now->lchild) q.push(now->lchild);
        if (now->rchild) q.push(now->rchild);
    }
}

int main(){
    cin >> n;
    hou.resize(n);
    mid.resize(n);
    for (int i = 0; i < n; i++){
        cin >> hou[i];
    }
    for (int i = 0; i < n; i++){
        cin >> mid[i];
    }
    node * root = nullptr;
    buildtree(root, 0, n - 1, 0, n - 1);
    findlevel(root);
    for (int i = 0; i < (int)level.size(); i++){
        if (i != 0) cout << " ";
        cout << level[i];
    }
    cout << endl;

    return 0;
}