思路:根据前序遍历依次访问对应的中序遍历的节点,分为左子树和右子树创建。
#include<iostream>
#include<stdlib.h>
using namespace std;
struct BinaryTreeNode
{
BinaryTreeNode(int _value)
:m_nValue(_value)
,m_pLeft(NULL)
,m_pRight(NULL)
{}
int m_nValue;
struct BinaryTreeNode* m_pLeft;
struct BinaryTreeNode* m_pRight;
};
BinaryTreeNode* Buildtree(int* pre,int* mid,int n)
{
if(n==0)
{
return NULL;
}
int num=pre[0];
BinaryTreeNode* head=new BinaryTreeNode(num);
int i=0;
while(i<n&&mid[i]!=num) //求左子树的长度
{
i++;
}
int left_len=i; //左子树的长度
int right_len=n-1-i; //右子树的长度
if(left_len>0) //构建左子树
{
head->m_pLeft=Buildtree(&pre[1],&mid[0],left_len);
}
if(right_len>0) //构建右子树
{
head->m_pRight=Buildtree(&pre[left_len+1],&mid[left_len+1],right_len);
}
return head;
}
void PreOrder(BinaryTreeNode* head)
{
if(head==NULL)
{
return;
}
cout<<head->m_nValue<<"->";
PreOrder(head->m_pLeft);
PreOrder(head->m_pRight);
}
void MidOrder(BinaryTreeNode* head)
{
if(head==NULL)
{
return;
}
MidOrder(head->m_pLeft);
cout<<head->m_nValue<<"->";
MidOrder(head->m_pRight);
}
int main()
{
int pre[]={1,2,4,7,3,5,6,8};
int mid[]={4,7,2,1,5,3,8,6};
BinaryTreeNode* head=Buildtree(pre,mid,8);
PreOrder(head);
cout<<endl;
MidOrder(head);
system("pause");
return 0;
}
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