标签:
一般我解题都是用的头递归或者尾递归。第一次
见识到了中递归,相当于把递归当成了一个循环体,用引用来作为变量,每一个递归中改动,须要非常强大的想象力。把整个递归树在脑子里想清楚。空间和时间都为O(n):
TreeNode *sortedListToBST(ListNode *head)
{
vector<TreeNode*> treeNodes;
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;
}
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;
}解法引用:http://www.bwscitech.com/a/jishuzixun/javayuyan/2013/0930/15822.htmlConvert Sorted List to Binary Search Tree
标签:
原文地址:http://www.cnblogs.com/yxwkf/p/5205180.html