PAT.A1020 Tree Traversals

Back to Contents

Title

Given the sequence of post-order traversal and middle-order traversal of a tree, find the hierarchical traversal sequence.

Sample (can be copied)

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

//output
4 1 6 3 5 7 2

important point

1. The order of several traversals. First order traversal: center-left-right. Mid-order traversal: left-center-right. Post-order traversal: left-right-center.
2. It is recommended that the reader manually simulate the create function to build the tree given in the example.

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

struct  node{
	int data;
	node* lc;
	node* rc;
};
const int maxn=50;
int per[maxn],post[maxn],in[maxn],n,num=0;
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;//新的根节点
	for(k=inl;k<=inr;k++){
		if(in[k]==post[postr])break;
	}
	int numleft=k-inl;//左子树数量
	root->lc=create(postl,postl+numleft-1,inl,k-1);
	root->rc=create(postl+numleft,postr-1,k+1,inr);
	return root;
}
void BFS(node* root){
	queue<node*> q;//层次遍历使用queue
	q.push(root);
	while(!q.empty()){
		node* now=q.front();
		q.pop();
		printf("%d",now->data);
		num++;
		if(num<n)printf(" ");
		if(now->lc!=NULL)q.push(now->lc);
		if(now->rc!=NULL)q.push(now->rc);
	}
}
int main(){
	cin>>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;
}