109. Convert Sorted List to Binary Search Tree

O(NlogN)

先找中点，然后分别left right

class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
            return null;
        }

        if (head.next == null) {
            return new TreeNode(head.val);
        }
        
        // 找链表中点
        ListNode slow = head, fast = head;
        ListNode preSlow = null; // to cut the list before mid
        while (fast != null && fast.next != null) {
            preSlow = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        // slow是中点
        ListNode mid = slow;
        // cut the list before mid!!
        preSlow.next = null;

        TreeNode root = new TreeNode(mid.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(mid.next);

        return root;
    }
}

O(logN)

• leftmost element in the inorder traversal has to be the head of our given linked list.
• Similarly, the next element in the inorder traversal will be the second element in the linked list and so on.
• so we can do sth like

 ➔ function formBst(start, end)
➔      mid = (start + end) / 2
 ➔      formBst(start, mid - 1)
 ➔
 ➔      TreeNode(head.val)
 ➔      head = head.next
 ➔       
 ➔      formBst(mid + 1, end)
 ➔

模拟中序遍历

class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        ListNode cur = head;
        int len = 0;
        while (cur != null) {
            cur = cur.next;
            len++;
        }

        curHead = head;

        return buildBST(0, len - 1);


    }

    ListNode curHead;

    public TreeNode buildBST(int start, int end) {

        if (start > end) {
            return null;
        }

        int mid = (start + end) / 2;
        
        // inorder，先遍历left
        TreeNode left = buildBST(start, mid - 1);

        // inorder，遍历root
        TreeNode root = new TreeNode(curHead.val);
        root.left = left;

        curHead = curHead.next;

        // inorder，后遍历right
        root.right = buildBST(mid + 1, end);

        return root;
        
    }
}