BST ： sorted list (array) to BST

Convert Sorted Array to Binary Search Tree

class Solution {
public:
    TreeNode* sortedArrayToBSTR(vector<int>& num, int start,int end)
{
	if(start > end)
		return NULL;
	int mid = (start+end)/2;
	TreeNode* r = new TreeNode(num[mid]);
	r->left = sortedArrayToBSTR(num, start, mid-1);
	r->right = sortedArrayToBSTR(num, mid+1, end);
	return r;
}


TreeNode *sortedArrayToBST(vector<int> &num)
{
	TreeNode* root = sortedArrayToBSTR(num,0, num.size()-1);
	return root;
}
};

Convert Sorted List to Binary Search Tree

 

看看自己的代码再看看别人的觉得自己做的好傻，直接传递长度就好了

class Solution {
public:
    
TreeNode* ListToBST(ListNode* start,ListNode* end)
{
	if(start == end)
		return NULL;
	ListNode* p = start;
	ListNode* q = start;
	while(q!=end && q->next != end)
	{
		q = q->next->next;
		p = p->next;
	}
	TreeNode* node = new TreeNode(p->val);
	node->left = ListToBST(start,p);
	node->right = ListToBST(p->next,end);//注意这里迭代就是end而不是NULL
	return node;
}

TreeNode* sortedListToBST(ListNode* head)
{
	if(head == NULL)
		return NULL;
	
	ListNode* p = head;
	ListNode* q = head;
	while(q!=NULL && q->next != NULL)
	{
		q = q->next->next;
		p = p->next;
	}
	TreeNode* root = new TreeNode(p->val);
	root->left = ListToBST(head,p);
	root->right = ListToBST(p->next,NULL);
	return root;
}
};

这是别人的代码

深度递归，直到底部再依次返回,等回去看明白为什么用*&，用*就不对，因为*&表示是引用指针，如果只是*那么就是值传递，形参是实参的副本，如果*&那么不需要复制，形参只是实参的别名，在递归中对指针的更改会被保存，不然返回到上次递归调用的地方，cur的值就是该环境下的值，递归中的值不被保存。

class Solution {
public:
    
TreeNode* ListToBST(ListNode* &cur,int start,int end)//重点是这里，cur前面还需要有引用符号不然出错
{
	if(start > end)
		return NULL;
	int mid = (start +end)/2;
	TreeNode* left = ListToBST(cur,start,mid-1);
	TreeNode* node = new TreeNode(cur->val);
	cur= cur->next;
	TreeNode* right = ListToBST(cur,mid+1,end);
	node->left = left;
	node->right = right;
	return node;
}

TreeNode* sortedListToBST(ListNode* head)
{
	if(head == NULL)
		return NULL;
	
	ListNode* p = head;
	int c = 0;
	while(p!=NULL)
	{
		p = p->next;
		++c;
	}
	ListToBST(head,0,c-1);
}
};