LeetCode Convert Sorted Array to Binary Search Tree

LeetCode Convert Sorted Array to Binary Search Tree

Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

一看题目，就感觉需要用二分法来做，因为数组是排序的，对应生成的树也要是二分搜索树。

具体分许见代码注释

//flag=1表示若此次调用成功，则是root的左子树接点；flag=2表示若此次调用成功，则是root的右子树节点
void ArrayToBST(TreeNode* root, vector<int>& num, int left, int right, int flag)
{
	//left==num.size()/2 || right==num.size()，由于中间节点在程序最开始的时候已经创建，因此这里需要排除，免得重复
	if (left > right || left==num.size()/2 || right==num.size())
		return;

	//创建节点，并挂在root下
	TreeNode* pNode = NULL;
	if (left == right)
	{		
		pNode = new TreeNode(num[left]);
		pNode->left = pNode->right = NULL;
		if (flag == 1)
			root->left = pNode;
		else
			root->right = pNode;
		
		return;
	}
	
	int mid = (left + right) / 2;
	pNode = new TreeNode(num[mid]);
	pNode->left = pNode->right = NULL;
	if (flag == 1)
		root->left = pNode;
	else
		root->right = pNode;

	//构建该节点的左子树
	ArrayToBST(pNode, num, left, mid-1, 1);
	//构建该节点的右子树
	ArrayToBST(pNode, num, mid+1, right, 2);
}

TreeNode *sortedArrayToBST(vector<int> &num)
{
	if (num.size() == 0)
		return NULL;
	
	int n = num.size();
	TreeNode* root = new TreeNode(num[n/2]);
	root->left = root->right = NULL;

	//构建根节点的左子树
	ArrayToBST(root, num, 0, n/2, 1);
	//构建根节点的右子树
	ArrayToBST(root, num, n/2+1, n-1, 2);

	return root;
}

以上代码不够简洁，经过网上的参考，优化之后的代码

TreeNode* buildTree(vector<int> &num, int left, int right)
{
	if (left > right)
		return NULL;

	int mid = left + right / 2;
	TreeNode* leftNode = buildTree(num, left, mid);
	TreeNode* rightNode = buildTree(num, mid + 1, right);

	TreeNode* node = new TreeNode(num[mid]);
	node->left = leftNode;
	node->right = rightNode;

	return node;
}


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