1.两颗二叉树A和B,判断B是不是A的子结构
/*
思路:1.在树A中找到和B的根节点一样的节点R
2.判断树A以R为根节点的子树是不是包含和树B一样的结构
*/
bool DoesTree1HaveTree2(BiTreeNode*T1,BiTreeNode*T2);
bool HasSubtree(BiTreeNode*T1,BiTreeNode*T2)
{
bool result=false;
if(T1&&T2)
{
if(T1->data==T2->data)
result=DoesTree1HaveTree2(T1,T2);
if(!result)
result=HasSubtree(T1->lchild,T2);
if(!result)
result=HasSubtree(T1->rchild,T2);
}
return result;
}
//T1是否包含T2
bool DoesTree1HaveTree2(BiTreeNode*T1,BiTreeNode*T2)
{
if(T2==NULL)
return true;
if(T1==NULL)
return false;
if(T1->data !=T2->data)
return false;
return DoesTree1HaveTree2(T1->lchild,T2->lchild)&&DoesTree1HaveTree2(T1->rchild,T2->rchild);
}
2.二叉树的镜像
void MirrorRecusively(BiTreeNode*T)
{
if((T==NULL)||(T->lchild==NULL&&T->rchild==NULL))
return;
BiTreeNode *temp=T->lchild;
T->lchild=T->rchild;
T->rchild=temp;
if(T->lchild)
MirrorRecusively(T->lchild);
if(T->rchild)
MirrorRecusively(T->rchild);
}
3.输入一个整型数组,判断该数组是不是某二叉搜索树的后序遍历
bool VerifySquenceOfBST(int sequence[],int length)
{
if(sequence==NULL||length<=0)
return false;
int root=sequence[length-1];
//左子树的节点小于根节点
int i=0;
for(;i<length-1;++i)
{
if(sequence[i]>root)
break;
}
//右子树的节点大于根节点
int j=i;
for(;j<length-1;++j)
{
if(sequence[j]<root)
return false;
}
bool left=true;
if(i>0)
left=VerifySquenceOfBST(sequence,i);
bool right=true;
if(i<length-1)
right=VerifySquenceOfBST(sequence+i,length-i-1);
return (left&&right);
}