PAT (Advanced Level) Practice-1020 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

#define maxn 35
int n;

int postorder[maxn], inorder[maxn];

typedef struct node{
	int value;
	node *left, *right;
	node(int v=0, node *left=NULL, node *right=NULL):value(v),left(left),right(right){}
}*root;

void build(root tree, int left, int right, int post){		//left,right是中序遍历的边界，post是元素在后序遍历中的序号
	if(tree==NULL)
		return ;
	int mid=0;
	for(int i=left; i<=right; ++i){
		if(inorder[i]==tree->value){
			mid = i;
			break;
		}
	}
	int rlen=right-mid;
	int llen=mid-left;
	if(llen>0){
		tree->left = new node(postorder[post-rlen-1]);			
		build(tree->left, left, mid-1, post-rlen-1);
	}
	else{
		tree->left=NULL;
	}
	if(rlen>0){
		tree->right = new node(postorder[post-1]);
		build(tree->right, mid+1, right, post-1);
	}
	else{
		tree->right=NULL;
	}
}


void detree(root tree){
	if(tree==NULL)
		return ;
	detree(tree->left);
	detree(tree->right);
	delete(tree);
}

int main(){
	queue<root> q;
	while(cin >> n){

		for(int i=1; i<=n; ++i){
			cin >> postorder[i];
		}
		for(int i=1; i<=n; ++i){
			cin >> inorder[i];
		}
		root tree = new node(postorder[n]);
		build(tree, 1, n, n);
		vector<int> v;
		q.push(tree);
		while(!q.empty()){
			root rt = q.front();
			if(rt->left!=NULL)
				q.push(rt->left);
			if(rt->right!=NULL)
				q.push(rt->right);
			q.pop();
			v.push_back(rt->value);
		}

		cout << v[0];
		for(int i=1; i<v.size(); ++i){
			cout << ' ' << v[i];
		}
		cout << endl;
		detree(tree);
	}
	return 0;
}