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Convert Sorted List to Binary Search Tree LeetCode Solution

Problem – Convert Sorted List to Binary Search Tree LeetCode Solution

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

``````Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.
``````

Example 2:

``````Input: head = []
Output: []
``````

Constraints:

• The number of nodes in `head` is in the range `[0, 2 * 104]`.
• `-105 <= Node.val <= 105`

Convert Sorted List to Binary Search Tree LeetCode Solution in Java

``````public class Solution {
}
public TreeNode toBST(ListNode head, ListNode tail){

while(fast!=tail&&fast.next!=tail){
fast = fast.next.next;
slow = slow.next;
}
}
``````

Convert Sorted List to Binary Search Tree LeetCode Solution in C++

``````class Solution {
public:
{
}

private:
{
return NULL;
if( head->next == tail )    //
{
TreeNode *root = new TreeNode( head->val );
return root;
}
while( temp != tail && temp->next != tail )    // 寻找中间节点
{
mid = mid->next;
temp = temp->next->next;
}
TreeNode *root = new TreeNode( mid->val );
root->left = sortedListToBST( head, mid );
root->right = sortedListToBST( mid->next, tail );
return root;
}
};
``````

Convert Sorted List to Binary Search Tree LeetCode Solution in Python

``````def sortedListToBST(self, head):
return
# here we get the middle point,
# even case, like '1234', slow points to '2',
# '3' is root, '12' belongs to left, '4' is right
# odd case, like '12345', slow points to '2', '12'
# belongs to left, '3' is root, '45' belongs to right
while fast and fast.next:
fast = fast.next.next
slow = slow.next
# tmp points to root
tmp = slow.next
# cut down the left child
slow.next = None
root = TreeNode(tmp.val)
root.right = self.sortedListToBST(tmp.next)
return root
``````
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