Convert Sorted List to Binary Search Tree

Total Accepted: 12283 Total
Submissions: 45910My Submissions

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

总结:这个是网上找的,还不明确为什么是以下这个样子,去掉&,就出错

ListNode *&node
TreeNode *sortedListToBST(ListNode *head)
{
int len = 0;
ListNode * node = head;
while (node != NULL)
{
node = node->next;
len++;
}
return buildTree(head, 0, len-1);
} TreeNode *buildTree(ListNode *&node, int start, int end)
{
if (start > end) return NULL;
int mid = start + (end - start)/2;
TreeNode *left = buildTree(node, start, mid-1);
TreeNode *root = new TreeNode(node->val);
root->left = left;
node = node->next;
root->right = buildTree(node, mid+1, end);
return root;
}

总结:这个開始的时候有点问题,改动了一下

TreeNode *sortedListToBST(ListNode *head)
{
vector<TreeNode*> treeNodes;
if(head == NULL) return NULL;
while (head != NULL)
{
TreeNode *node = new TreeNode(head->val);
treeNodes.push_back(node);
head = head->next;
}
return genBST(0, treeNodes.size()-1, treeNodes);
} TreeNode* genBST(int start, int end, vector<TreeNode*> &treeNodes)
{
if (start == end) return treeNodes[start];
else if (start+1 == end)
{
treeNodes[start]->right = treeNodes[end];
return treeNodes[start];
} int mid = (start+end)/2;
TreeNode* root = treeNodes[mid];
root->left = genBST(start, mid-1, treeNodes);
root->right = genBST(mid+1, end, treeNodes);
return root;
}

总结:上面这个代码跟我參考得到的思想一致,可是我的是time limit

TreeNode *sortedListToBST(ListNode *head) {
int len = 0;
ListNode *node = head;
while(node)
{
len++;
node = node->next;
}
return BuildTree(head,1,len);
} TreeNode *BuildTree(ListNode *node,int start,int end)
{
if(start > end) return NULL;
int mid = start + (end-start)/2;
ListNode *mi_node = node;
for(int i=1;i<mid;i++)
{
mi_node = mi_node->next;
} TreeNode *root = new TreeNode(mi_node->val);
root->left = BuildTree(node,start,mid-1);
root->right = BuildTree(node,mid+1,end);
return root;
}

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