1020 Tree Traversals

--write by zhuwx 2019-08-04 17:20:46 +0800 CST

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  (), 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


#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<queue>
using namespace std;
const int maxn=50;
struct node{
int data;
node* lchild;
node* rchild;
};
int pre[maxn];//先序
int in[maxn];//中序
int post[maxn];//后续
int n;

node* create(int postl,int postr,int inl,int inr){
if(postl>postr){
return NULL;
}
node* root=new node;
root->data=post[postr];
int k;
for(k=inl;k<inr;k++){
//寻找根节点
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!=NULL)
q.push(now->lchild);
if(now->rchild!=NULL)
q.push(now->rchild);
}
}

int main(){
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);
system("pause");
return 0;
}