腾讯面试题目--非递归遍历二叉树

代码如下：

// CreateBinaryTree1.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <stack>
using namespace std;

typedef struct node
{
	char data;
	struct node *lchild;
	struct node *rchild;
} treenode;

treenode* CreateBinTree()
{
	char x;
	cin>>x;
	treenode* temp;
	if ('$'==x)
	{
		temp=NULL;
	}
	else
	{
		temp = new treenode;
		temp->data = x;
		temp->lchild=CreateBinTree();
		temp->rchild=CreateBinTree();
	}
	return temp;
}
//前序遍历
void preorder(treenode* p)
{
	if(p!=NULL)
	{
		cout<<p->data<<" ";
		preorder(p->lchild);
		preorder(p->rchild);
	}
}
//中序遍历
void midorder(treenode* p)
{
	if(p!=NULL)
	{
		midorder(p->lchild);
		cout<<p->data<<" ";
		midorder(p->rchild);
	}
}

//后序遍历
void postorder(treenode* p )
{
	if(p!=NULL)
	{
		postorder(p->lchild);
		postorder(p->rchild);
		cout<<p->data<<" ";
	}
}

//非递归先根遍历
void PreOrder2(treenode *root)
{
	stack<treenode*>T;
	T.push(root);
	treenode *p=new treenode;
	while(!T.empty())
	{
		p=T.top();
		T.pop();
		cout<<p->data<<" ";
		if(p->rchild)
		{
			T.push(p->rchild);
		}
		if(p->lchild)
		{
			T.push(p->lchild);
		}
	}
	cout<<endl;
}
//非递归中序遍历
void midorder2(treenode *root)
{
	stack<treenode*>T;
	treenode *p=root;
	while((NULL!=p)||!T.empty())
	{
		if(NULL!=p)
		{
			T.push(p);
			p=p->lchild;
		}
		else
		{
			p=T.top();
			T.pop();
			cout<<p->data<<" ";
			p=p->rchild;
		}
	}
	cout<<endl;
}
//非递归后序遍历
void postOrder2(treenode *root)
{
	stack<treenode*>T;
	treenode *p=root;
	treenode *pre=NULL;
	while((NULL!=p)||!T.empty())
	{
		while(p)
		{
			T.push(p);
			p=p->lchild;
		}
		p=T.top();
		//如果p没有右孩子或者其右孩子刚刚被访问过
		if(p->rchild == NULL || p->rchild == pre)
		{
			cout<<p->data<<" ";
			T.pop();
			pre=p;
			p=NULL;
		}
		else
		{
			p=p->rchild;
		}
	}
	cout<<endl;
}

int _tmain(int argc, _TCHAR* argv[])
{
	cout<<"请以先根遍历输入一序列，$表示空:\n";
	treenode *root  = CreateBinTree();
	cout<<"前序遍历: ";
	preorder(root);
	cout<<endl;
	cout<<"中序遍历：";
	midorder(root);
	cout<<endl;
	cout<<"后序遍历：";
	postorder(root);
	cout<<endl;
	cout<<"非递归前序遍历:";
	PreOrder2(root);
	cout<<endl;
	cout<<"非递归中序遍历:";
	midorder2(root);
	cout<<endl;
	cout<<"非递归后序遍历:";
	postOrder2(root);
	cout<<endl;
	system("pause");
	return 0;
}

结果如下：