PAT 甲级 1127 ZigZagging on a Tree

发布时间：2020-12-14 05:18:16 所属栏目：大数据来源：网络整理

导读：https://pintia.cn/problem-sets/994805342720868352/problems/994805349394006016 ? Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder trav

https://pintia.cn/problem-sets/994805342720868352/problems/994805349394006016

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However,if you think the problem is too simple,then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is,starting from the root,print the numbers level-by-level,alternating between left to right and right to left. For example,for the following tree you must output: 1 11 5 8 17 12 20 15.

Input Specification:

Each input file contains one test case. For each case,the first line gives a positive integer N ( $\leq30),the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.$

Output Specification:

For each test case,print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

代码：

#include <bits/stdc++.h>
using namespace std;

int N,root;
vector<int> in,post;
vector<int> ans[35];
int tree[35][2];

struct Node{
    int index;
    int depth;
};

void dfs(int &index,int ileft,int iright,int pleft,int pright) {
    if(ileft > iright) return ;
    index = pright;
    int i = 0;
    while(in[i] != post[pright]) i ++;
    dfs(tree[index][0],ileft,i - 1,pleft,pleft + i - ileft - 1);
    dfs(tree[index][1],i + 1,iright,pleft + i - ileft,pright - 1);
}

void bfs() {
    queue<Node> q;
    q.push(Node{root,0});
    while(!q.empty()) {
        Node temp = q.front();
        q.pop();
        ans[temp.depth].push_back(post[temp.index]);
        if(tree[temp.index][0])
            q.push(Node{tree[temp.index][0],temp.depth + 1});
        if(tree[temp.index][1])
            q.push(Node{tree[temp.index][1],temp.depth + 1});
    }
}

int main() {
    scanf("%d",&N);
    in.resize(N + 1),post.resize(N + 1);
    for(int i = 1; i <= N; i ++)
        scanf("%d",&in[i]);
    for(int i = 1; i <= N; i ++)
        scanf("%d",&post[i]);

    dfs(root,1,N,N);
    bfs();

    printf("%d",ans[0][0]);
    for(int i = 1; i < 35; i ++) {
        if(i % 2) {
            for(int j = 0; j < ans[i].size(); j ++)
                printf(" %d",ans[i][j]);
        } else {
            for(int j = ans[i].size() - 1; j >= 0; j --)
                printf(" %d",ans[i][j]);
        }
    }

    return 0;
}

　　可能敲了一万遍才敲对吧今天是被 Tizzy T 洗脑的一天

（编辑：李大同）

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