1020 Tree Traversals (25 分)(⼆叉树的遍历,后序中序转层序)
作者:互联网
Description
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 (<=30), 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
生词
英文 | 解释 |
---|---|
distinct | 不同的 |
traversal | 遍历 |
corresponding | 相应的 |
题目大意:
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
分析:
与后序中序转换为前序的代码相仿(无须构造二叉树再进行广度优先搜索~),只不过加一个变量index,表示当前的根结点在二叉树中所对应的下标(从0开始),所以进行一次输出先序的递归过程中,就可以把根结点下标index及所对应的值存储在map<int, int> level中,map是有序的会根据index从小到大自动排序,这样递归完成后level中的值就是层序遍历的顺序~~
如果你不知道如何将后序和中序转换为先序,请看-> https://www.liuchuo.net/archives/2090
原文链接:https://blog.csdn.net/liuchuo/article/details/52137796
题解
终于把这篇在11.04就编辑的博文抄完了。。。
学废了之后发现也不是很难,可能忘太多了觉得太陌生了,有畏惧感呜呜
总而言之,还是要迎难而上!
#include <bits/stdc++.h>
const int maxn=50;
using namespace std;
struct node
{
int data;
node* lchild;
node* rchild;
};
int pre[maxn],in[maxn],post[maxn];
int n;
node* create(int postL,int postR,int inL,int inR)
{
if(postL>postR) return nullptr;
node* root=new node;
root->data=post[postR];
int k;
for(k=inL;k<=inR;k++){
if(in[k]==post[postR]) break;
}
int numLeft=k-inL;
root->lchild=create(postL,postL+numLeft-1,inL,k-1);
root->rchild=create(postL+numLeft,postR-1,k+1,inR);
return root;
}
int num=0;
void BFS(node* root){
queue<node*> q;
q.push(root);
while(!q.empty()){
node* now=q.front();
q.pop();
printf("%d",now->data);
num++;
if(num<n)printf(" ");
if(now->lchild!=nullptr) q.push(now->lchild);
if(now->rchild!=nullptr) q.push(now->rchild);
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&post[i]);
}
for(int i=0;i<n;i++){
scanf("%d",&in[i]);
}
node* root=create(0,n-1,0,n-1);
BFS(root);
return 0;
}
柳神代码(码住,以后看!)
#include <cstdio>
#include <vector>
#include <map>
using namespace std;
vector<int> post, in;
map<int, int> level;
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++;
level[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);
auto it = level.begin();
printf("%d", it->second);
while(++it != level.end()) printf(" %d", it->second);
return 0;
}
标签:node,25,1020,level,int,index,中序,post,root 来源: https://www.cnblogs.com/moonlight1999/p/15507553.html