《编程之美》3.10分层遍历二叉树

#include<iostream>
#include<vector>
using namespace std;

const int Max=30;
class Node
{
public:
Node(int d):data(d),left(NULL),right(NULL){}
int data;
Node *left;
Node *right;
};

template<class T>
class queue
{
public:
queue(int i=Max):front(0),rear(0){Q=new T[Max];}
void EnQueue(Node *p){Q[rear++] = p;}
T DeQueue(){return Q[front++];}
int front;
int rear;
T *Q;
};

int PrintNodeAtLevel(Node *root,int level)//递归
{
if(!root || level<0)
return 0;
if(level==0)
{
cout<<root->data<<" ";
return 1;
}
PrintNodeAtLevel(root->left,level-1);
PrintNodeAtLevel(root->right,level-1);
}
void PrintNode(Node *root)
{
for(int level=0; ;level++)
{
if(!PrintNodeAtLevel(root,level))
break;
cout<<endl;
}
}

void PrintNodeByLevel(Node *root)//非递归
{
if(root==NULL)
return;
vector<Node *> vec;
vec.push_back(root);
int cur=0;
int last=1;
while(cur<vec.size())
{
last=vec.size();
while(cur<last)
{
cout<<vec[cur]->data<<" ";
if(vec[cur]->left)
vec.push_back(vec[cur]->left);
if(vec[cur]->right)
vec.push_back(vec[cur]->right);
cur++;
}
cout<<endl;
}
}

void PrintNodeByBFS(Node *root) //BFS
{
if(root==NULL)
return;
queue<Node *> que;
que.EnQueue(root);
Node *q;
while(que.front<que.rear)
{
int	last=que.rear;
while(que.front<last)
{
q=que.DeQueue();
cout<<q->data<<" ";
if(q->left)
que.EnQueue(q->left);
if(q->right)
que.EnQueue(q->right);
}
cout<<endl;
}
}
//扩展问题
/*
*依然是按层遍历二叉树，只是要求从下往上访问，并且每一层中结点的访问顺序为从右向左

*分析：只要层与层之间加入哑元素（NULL），然后逆序输出队列Q即可

*/
void PrintNodeDown2UpRight2Left(Node *root)
{
if(root==NULL)
return;
queue<Node *> que;
que.EnQueue(root);
que.EnQueue(NULL);
Node *q;
while(que.front<que.rear-1)
{
int last=que.rear-1;
while(que.front<last)
{
q=que.DeQueue();
if(q->left!=NULL)
que.EnQueue(q->left);
if(q->right!=NULL)
que.EnQueue(q->right);
}
que.EnQueue(NULL);//如果左右子树为空又多加入一个NULL
que.front=last+1;
}

for(int i=que.rear-3;0<=i;i--)//逆序打印
{
if(que.Q[i]!=NULL)
cout<<que.Q[i]->data<<" ";
else
cout<<endl;
}
cout<<endl;//只是为了排版

}

int main()
{
Node *p1 = new Node(1);
Node *p2 = new Node(2);
Node *p3 = new Node(3);
Node *p4 = new Node(4);
Node *p5 = new Node(5);
Node *p6 = new Node(6);
Node *p7 = new Node(7);
Node *p8 = new Node(8);
p1->left = p2;
p1->right = p3;
p2->left = p4;
p2->right = p5;
p3->right = p6;
p5->left = p7;
p5->right = p8;
Node *root = p1;
//PrintNode(root);
//PrintNodeByLevel(root);
//PrintNodeByBFS(root);
PrintNodeDown2UpRight2Left(root);
return 0;
}