109. Convert Sorted List to Binary Search Tree

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

Have you met this question in a real interview? 

Yes

Example

               21->2->3  =>   / \             1   3

java

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /*     * @param head: The first node of linked list.     * @return: a tree node     */    private ListNode current;    public TreeNode sortedListToBST(ListNode head) {        // write your code here        if (head == null) {            return null;        }        current = head;        int size = getNum(head);        TreeNode node = toTree(size);        return node;    }    private TreeNode toTree(int size) {        if (size <= 0) {            return null;        }        TreeNode left = toTree(size / 2);        TreeNode root = new TreeNode(current.val);        current = current.next;        TreeNode right = toTree(size - 1 - size / 2);        root.left = left;        root.right = right;        return root;    }    private int getNum(ListNode head) {        int num = 0;        while (head != null) {            num++;            head = head.next;        }        return num;    } }

python

"""Definition of ListNodeclass ListNode(object):    def __init__(self, val, next=None):        self.val = val        self.next = nextDefinition of TreeNode:class TreeNode:    def __init__(self, val):        self.val = val        self.left, self.right = None, None"""class Solution:    """    @param: head: The first node of linked list.    @return: a tree node    """    node = ListNode(0)    def sortedListToBST(self, head):        # write your code here        if head == None:            return None        self.node = head        size = self.getSize(head)        return self.toTree(size)            def getSize(self, head):        num = 0        while head != None:            head = head.next            num += 1        return num    def toTree(self, size):        if size <= 0:            return None        left = self.toTree(size / 2)        root = TreeNode(self.node.val)        self.node = self.node.next        right = self.toTree(size - 1 - size / 2)        root.left = left        root.right = right        return root