HDU1622(Trees on the level)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1622

Trees on the level

Problem Description
Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers such as Thinking Machines' CM-5 are based on fat trees. Quad- and octal-trees are fundamental to many algorithms in computer graphics.

This problem involves building and traversing binary trees.
Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes.

In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.

For example, a level order traversal of the tree


is: 5, 4, 8, 11, 13, 4, 7, 2, 1.

In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L's and R's where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.




Input
The input is a sequence of binary trees specified as described above. Each tree in a sequence consists of several pairs (n,s) as described above separated by whitespace. The last entry in each tree is (). No whitespace appears between left and right parentheses.

All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.




Output
For each completely specified binary tree in the input file, the level order traversal of that tree should be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a node is given a value more than once, then the string ``not complete'' should be printed


Sample Input
(11,LL) (7,LLL) (8,R)
(5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
(3,L) (4,R) ()


Sample Output
5 4 8 11 13 4 7 2 1
not complete

题意:建立二叉树,对二叉树进行层次遍历

//思路:建立树,然后广搜遍历
//wa了几次
//此题有以下条件
//1.有子节点的父节点值不能为空
//2.输入() 输出not complete
//3.当输入(1,) (2,L) (2,L) (3,R)节点值重复时,要等输入()后才能输出not complete,切记不能在输完(2,L) (2,L)就输出not complete,我就在这里wa了几次，汗
#include <iostream>
#include <queue>
using namespace std;

typedef struct tNode{
	int data;
	struct tNode *lchild, *rchild;
	tNode()
	{
		lchild = rchild = NULL;
		data = -9999999;
	}
}*Node;
bool flag;
class Tree{
private:
	Node root;
public:
	Tree()
	{
		root = new tNode();
	}
	void insert(int data, char path[]); //增加节点
	void print(int d); //输出
	void destory(); //释放内存
	bool isBitTree(); //判断是否为二叉树
};

void Tree::insert(int data, char path[])
{
	int len = strlen(path);
	int i;
	Node cur = root;
	for(i = 0; i < len - 1; i++)
	{
		if(path[i] == 'L')
		{
			if(cur->lchild == NULL)
				cur->lchild = new tNode();
			cur = cur->lchild;
		}else if(path[i] == 'R')
		{
			if(cur->rchild == NULL)
				cur->rchild = new tNode();
			cur = cur->rchild;
		}
	}
	if(cur->data == -9999999)
		cur->data = data;
	else
		flag = false;
}
bool Tree::isBitTree()
{
	queue<Node> q;
	q.push(root);
	while(!q.empty())
	{
		Node cur = q.front();
		q.pop();
		if(cur->data == -9999999)
			return false;
		if(cur->lchild)
			q.push(cur->lchild);
		if(cur->rchild)
			q.push(cur->rchild);
	}
	return true;
}
void Tree::destory()
{
	queue<Node> q;
	q.push(root);
	while(!q.empty())
	{
		Node cur = q.front();
		q.pop();
		if(cur->lchild)
			q.push(cur->lchild);
		if(cur->rchild)
			q.push(cur->rchild);
		delete cur;
	}
	root = new tNode();
}
void Tree::print(int d)
{
	queue<Node> q;
	q.push(root);
	int w = 0;
	while(!q.empty())
	{
		Node cur = q.front();
		q.pop();
		if(w == d - 1)
			cout<<cur->data;
		else
			cout<<cur->data<<" ";
		if(cur->lchild)
			q.push(cur->lchild);
		if(cur->rchild)
			q.push(cur->rchild);
		w++;
	}
	cout<<endl;
}
int main()
{
	char snode[600];
	int data, d = 0;
	Tree t;
	flag = true;
	while(cin>>snode)
	{
		if(strcmp(snode, "()") == 0)
		{
			if(t.isBitTree() && flag)
					t.print(d);
			else
				cout<<"not complete"<<endl;
			t.destory();
			d = 0;
			flag = true;
			continue;
		}
		sscanf(snode + 1, "%d", &data);
		t.insert(data, strchr(snode, ',') + 1);
		d++;
	}
	return 0;
}