PAT 1020 Tree Traversals [后序+中序建树] [层次遍历]

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

--------------------------------------这是题目和解题的分割线--------------------------------------

 模拟一下已知中序序列和XX序列求XX序列的思路即可。

#include<cstdio>
#include<queue>

using namespace std;

typedef struct Node
{
	int data;
	Node *left,*right;
}node;

int post[35],in[35];

//[postL,postR]后序区间，[inL,inR]中序区间 
node* create(int postL,int postR,int inL,int inR)
{
	//如果序列小于等于0，直接返回 
	if(postL>postR) return NULL;
	node* root = new node;
	root->data = post[postR]; //根结点（后序的根结点在最后一个） 
	int i;
	//在中序里找到根的位置 
	for(i=inL;i<=inR;i++)
		if(in[i]==root->data) break;
	//左子树的结点个数 
	int numLeft = i - inL;
	//左子树的后序区间和中序区间 
	root->left = create(postL,postL+numLeft-1,inL,inL+numLeft-1);
	//右子树的后序区间和中序区间 
	root->right = create(postL+numLeft,postR-1,i+1,inR);
	return root;
}

int n,count = 0;

void levelOrder(node *T)
{
	queue<node*> q;
	q.push(T);
	while(!q.empty())
	{	
		node* top = q.front();
		q.pop();
		printf("%d",top->data);
		if(count!=n-1) printf(" "); //末尾不能输出空格 
		count++;
		if(top->left) q.push(top->left);
		if(top->right) q.push(top->right);
	}
}

int main()
{
	int i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&post[i]);
	for(i=0;i<n;i++)
		scanf("%d",&in[i]);
	node* T = create(0,n-1,0,n-1);
	levelOrder(T);
	return 0;
}