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.

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:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

Solution

Approach 1: use 2 pointers to find root (at the middle of linked list), then recurse. Time: NlogN. Extra space: logN (stack).

Approach 2: convert to array, then recurse. Time: N. Extra space: logN (stack) and N (array).

Approach 3: Simulate what an inorder traversal does: recursively construct the left subtree, then use the current head as root, then recursively contruct the right subtree. Maintain the invariant that once a linkedlist node is used, the current head moves forward. Time: N. Extra space: logN (stack).

Code (Python)

Approach 3:

    def sortedListToBST(self, head: 'ListNode') -> 'TreeNode':
        if not head:
            return None
        self.head = head
        len_of_list = 0
        node = head
        while node:
            len_of_list += 1
            node = node.next
        return self._to_bst(0, len_of_list - 1)
    
    # the start, end and mid pointers are there to limit the base case (make sure the base case happens)
    def _to_bst(self, start, end):
        if start > end:
            return None
        mid = (start + end) // 2
        left = self._to_bst(start, mid - 1)
        root = TreeNode(self.head.val)
        self.head = self.head.next
        root.left = left
        root.right = self._to_bst(mid + 1, end)
        return root

Code (C++)

Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
// Recursion.
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if (head == NULL)
            return NULL;
        ListNode *slow = head;
        ListNode *fast = head;
        ListNode dummyHead(0);
        dummyHead.next = head;
        ListNode *prev = &dummyHead;
        while (fast && fast->next)
        {
            fast = fast->next->next;
            slow = slow->next;
            prev = prev->next;
        }
        TreeNode *root = new TreeNode(slow->val);
        prev->next = NULL;
        root->left = sortedListToBST(dummyHead.next);
        root->right = sortedListToBST(slow->next);
        return root;
    }
};

Approach 3:

// Inorder simulation.
class Solution {
private:
    ListNode *curr;
    TreeNode* formBst(int head, int tail)
    {
        if (head > tail)
            return NULL;
        int mid = head + (tail - head) / 2;
        TreeNode *left = formBst(head, mid - 1);
        TreeNode *root = new TreeNode(curr->val);
        curr = curr->next;
        root->left = left;
        root->right = formBst(mid + 1, tail);
        return root;
    }
public:
    TreeNode* sortedListToBST(ListNode* head) {
        // Get the length of the linked list.
        int len = 0;
        ListNode *p = head;
        while (p)
        {
            len++;
            p = p->next;
        }
        curr = head;
        return formBst(0, len - 1);
    }
};