Pat grade a A1020 tree Traversals

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

The question ： Give how many nodes the binary tree has , The next first sequence is the post order traversal sequence , The second is the middle order ergodic sequence , According to these two sequences, find the hierarchical traversal sequence .

Ideas : Construct a binary tree according to two traversal sequences , Then use sequence traversal （ Breadth first search -- The branch always accesses the node that can be reached directly first （ Put the left and right children of the node into the queue first ）） Find the hierarchical traversal sequence

Code ：
 

#include<iostream>
#include<queue>
using namespace std;
const int maxn=50;
int in[maxn],post[maxn];
int n;
struct node{
	int data;
	node* lchild;
	node* rchild;
};
// Construct binary tree according to sequence ; 
node* create(int postL,int postR,int inL,int inR){
	if(postL>postR)return NULL;// Recursive boundary ;
	node* root=new node;// Construct the root node ; 
	root->data=post[postR];
	int k;
	for(k=inL;k<=inR;k++){// Notice that this is inL To inR, Find the root node in the subtree , Not from 0 To inR, And because the incoming is 0 To n-1, So it can be equal to inR Of ; 
		if(in[k]==root->data){
			break;
		}
	}
	int leftnum=k-inL;// Confirm the left and right subtrees ;
	root->lchild=create(postL,postL+leftnum-1,inL,k-1);// Then enter the left subtree and repeat the operation , Bring two sequences into the function , Find the root node , Create a root node , Confirm the left and right subtrees , Until the subsequence is numbered （ When it is an empty tree ）, return NULL; 
	root->rchild=create(postL+leftnum,postR-1,k+1,inR);// Recursive ; 
	return root; 
}
// Breadth first search for hierarchical sequence ; 
void BFS(node* root){
	queue<node* > Q; 
	// Note that here is the pointer queue , Because the queue stores copies of binary tree nodes , If you want to modify the data of the binary tree , You need to use the pointer of this node , Go to the location of the node to modify , So this stores pointers ; 
	Q.push(root);
	int num=0;
	while(!Q.empty()){
		node* top=Q.front();
		Q.pop();
		printf("%d",top->data);
		num++;
		if(num<n)printf(" ");
		if(top->lchild!=NULL)Q.push(top->lchild);
		if(top->rchild!=NULL)Q.push(top->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);
	return 0;
}