1020 Tree Traversals (25 分)

1020 Tree Traversals (25 分)

 

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2


这道题我感觉像是在测你脑回路，挺有趣的，
这题我用bfs写的，先给你后序遍历（左右中）和中序遍历（左中右）。
也有看到一些别的题解，比如柳诺大神的题解，但是她的解法如果数据强点就过不去了。
所以后台数据确实比较水，但是用bfs的代码就不用担心数据量的问题了。
比如下面这组数据：

30
30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

这组数据她的代码绝对跑不了，因为2^30可是超了int，就算是vector也不可能开这么多，而且那样很浪费空间
用个bfs写出来就时间空间都解决了。

#include <bits/stdc++.h>
#define N 100
using namespace std;
int an[40], bn[40], n, cn[N],pos = 0;

struct Node{
    int root, start, end;
};
queue<Node> q;
void dfs(int root, int start, int end){
    Node node;
    node.root = root, node.start = start,node.end = end;
    q.push(node);
    while(!q.empty()){
        node = q.front();
        q.pop();
        if(node.start > node.end) 
            continue;
        cn[pos++] = an[node.root];
        int i;
        for(i = node.start; i < node.end; i++){
            if(bn[i] == an[node.root])
                break;
        }
        q.push({node.root-(node.end-i)-1, node.start, i-1});
        q.push({node.root-1, i+1, node.end}); 
    }
}

int main(){
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> an[i];
    }
    for(int i = 0; i < n; i++){
        cin >> bn[i];
    }
    memset(cn,-1,sizeof(cn));
    dfs(n-1, 0, n-1);
    for(int i = 0; i < pos; i++)
        printf("%d%c",cn[i],i==pos-1?'\n':' ');
    
    return 0;
}

 

这里给了一下柳神的代码（虽然是错的，但是可以过题，想法还是挺好的）：^v^(勿喷)

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct node {
    int index, value;
};
bool cmp(node a, node b) {
    return a.index < b.index;
}
vector<int> post, in;
vector<node> ans;
void pre(int root, int start, int end, int index) {
    if (start > end) return;
    int i = start;
    while (i < end && in[i] != post[root]) i++;
    ans.push_back({index, post[root]});
    pre(root - 1 - end + i, start, i - 1, 2 * index + 1);
    pre(root - 1, i + 1, end, 2 * index + 2);
}
int main() {
    int n;
    scanf("%d", &n);
    post.resize(n);
    in.resize(n);
    for (int i = 0; i < n; i++) scanf("%d", &post[i]);
    for (int i = 0; i < n; i++) scanf("%d", &in[i]);
    pre(n - 1, 0, n - 1, 0);
    sort(ans.begin(), ans.end(), cmp);
    for (int i = 0; i < ans.size(); i++) {
        if (i != 0) cout << " ";
        cout << ans[i].value;
    }
    return 0;
}