PAT甲級 1020

題目 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.

解析

已知二叉樹的後序遍歷、中序遍歷，求二叉樹的層序遍歷

代碼

#include<bits/stdc++.h> 

using namespace std;

struct node{
	int data;
	node *lchild,*rchild;
};

int n,ind=0;
int postorder[35], inorder[35],preorder[35];
node * res;

node* create(int i1,int i2, int p1, int p2){
	if(i1>i2 || p1>p2) return NULL;
	node *root = new node;
	root->data = postorder[p2];
	int k;
	for(k = i1; k < i2;k++)
		if(root->data == inorder[k])
			break; 
	
	root->lchild = create(i1, k-1, p1, p1+k-i1-1);
	root->rchild = create(k+1, i2, p2+k-i2, p2-1);
	return root;
}

void bfs(node *root){
	queue<node*> q;
	q.push(root);
	
	while(!q.empty()){
		node *root = q.front();
		q.pop();
		preorder[ind++] = root->data;
		
		if(root->lchild !=NULL) 
			q.push(root->lchild);
		if(root->rchild !=NULL) 
			q.push(root->rchild);
		
	}
	
} 

int main(){
	
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>postorder[i];
	for(int i=0;i<n;i++)
		cin>>inorder[i];
		
	res = create(0, n-1, 0, n-1);
	bfs(res);
	for(int i=0;i<n;i++){
		cout<<preorder[i];
		if(i!=n-1){
			printf(" ");
		}
	}
	
	return 0;
}