1.兩顆二叉樹A和B，判斷B是不是A的子結構

/*
	思路：1.在樹A中找到和B的根節點一樣的節點R
	     2.判斷樹A以R為根節點的子樹是不是包含和樹B一樣的結構 
*/ 
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);
		 	
}
 

2.二叉樹的映象

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.輸入一個整型陣列，判斷該陣列是不是某二叉搜尋樹的後序遍歷

bool VerifySquenceOfBST(int sequence[],int length)
{
	if(sequence==NULL||length<=0)
		return false;
	int root=sequence[length-1];
	
	//左子樹的節點小於根節點
	int i=0;
	for(;i<length-1;++i) 
	{
		if(sequence[i]>root)
			break;
	}
	
	//右子樹的節點大於根節點
	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);
}