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 sortedListToBST(ListNode head) {
    if(head==null) return null;
    return toBST(head,null);
}
public TreeNode toBST(ListNode head, ListNode tail){
    ListNode slow = head;
    ListNode fast = head;
    if(head==tail) return null;
    
    while(fast!=tail&&fast.next!=tail){
        fast = fast.next.next;
        slow = slow.next;
    }
    TreeNode thead = new TreeNode(slow.val);
    thead.left = toBST(head,slow);
    thead.right = toBST(slow.next,tail);
    return thead;
}

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

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head)
    {
    	return sortedListToBST( head, NULL );
    }
    
private:
    TreeNode *sortedListToBST(ListNode *head, ListNode *tail)
    {
    	if( head == tail )
    		return NULL;
    	if( head->next == tail )    // 
    	{	
    		TreeNode *root = new TreeNode( head->val );
    		return root;
    	}
    	ListNode *mid = head, *temp = head;
    	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):
    if not head:
        return 
    if not head.next:
        return TreeNode(head.val)
    # 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
    slow, fast = head, head.next.next
    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.left = self.sortedListToBST(head)
    root.right = self.sortedListToBST(tmp.next)
    return root
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