首页 > 代码库 > travel the binary tree by level 5 ( from down to top and from left to right every level )

travel the binary tree by level 5 ( from down to top and from left to right every level )

travel the binary tree by level 5 ( from down to top and from left to right every level )

个人信息:就读于燕大本科软件工程专业 目前大三;

本人博客:google搜索“cqs_2012”即可;

个人爱好:酷爱数据结构和算法,希望将来从事算法工作为人民作出自己的贡献;

博客内容:travel the binary tree by level 5 ( from down to top and from left to right every level )

博客时间:2014-5-4;

编程语言:C++ ;

编程坏境:Windows 7 专业版 x64;

编程工具:vs2008 32位编译器;

制图工具:office 2010 ppt;

硬件信息:7G-3 笔记本;


my words

Money can be stolen, but not knowledge. Studying benefits me forever.

problem( from beauty of programming )

travel the binary tree by level, and visit nodes form left to right every level.

make the function:  int TravelByLevel(node * root, int level)

eg: the binary tree follows


the 2 level nodes: 3,7,12,15

my solution

first action: make tree stored with a two-dimensional table like following


then getting nodes of every level becomes very easy. 

time cost O(n), n is the numbers node of the binary tree, and this algorithm is faster  than that algorithm with queue and stack which I proposed yesterday.

function follows( we assume the binary tree is not so big, or new more space just ok)

int _TravelByLevel(node * T,int level)
{
	cout<<"travel start level "<<level<<endl;
	// first action
	table * table_tree = new table[length];
	queue<std::pair<node *,int>> Q;

	if(T != NULL)
		Q.push(std::make_pair(T,0));

	// second action
	std::pair<node*,int> p;
	while(! Q.empty())
	{
		// first action: 2.1
		p = Q.front();
		Q.pop();
		((table_tree[p.second]).list)[( (table_tree[p.second]).num)++] = p.first;

		// second action: 2.2
		if((p.first)->left != NULL)
			Q.push(std::make_pair((p.first)->left,p.second+1));
		if((p.first)->right != NULL)
			Q.push(std::make_pair((p.first)->right,p.second+1));
	}
	
	// third action
	node * p1;
	
	if( level < length && table_tree[level].num > 0 )
	{
		for(int j =0;j<(table_tree[level]).num;j++)
		{
			p1 = ((table_tree[level]).list)[j];
			cout<<p1->data<<endl;
		}
		cout<<"travel over"<<endl;
		return 1;
	}
	cout<<"travel over"<<endl;
	return 0;
}

program run out:


my code

test.cpp

#include<iostream>
#include<queue>
#include<stack>
using namespace std;

const int length = 100;
//const int Longlength = 10000;



class node
{
public:
    int data ;
    node * left ;
    node * right ;
    node()
	{
        data = http://www.mamicode.com/0 ;>