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
看示例，看起来关键是“每个节点 的左右两个子树的高度差的绝对值不超过 1。”是不是平衡二叉树不重要？看到二叉树这种东西就心虚，先存个链接
这样似乎有个思路：快慢指针遍历链表，获取到中间位置。分别翻转前后两个链表，然后按照示例的方式构造二叉树？
参考了这里的实现，验证我的思路是(一半)正确的。 但我自己实现了半天也没出来- -||，还是个半成品。
Runtime: 1 ms, faster than 94.08% of Java online submissions for Convert Sorted List to Binary Search Tree.
Memory Usage: 38.3 MB, less than 100.00% of Java online submissions for Convert Sorted List to Binary Search Tree.
public TreeNode sortedListToBST(ListNode head) {
if (head == null) {
return null;
}
if (head.next == null) {
return new TreeNode(head.val);
}
ListNode slow = head, pre = null, fast = head;
//快慢指针走到中间：走到中间的判断条件——判断fast && fast.next
while (fast != null && fast.next != null) {
pre = slow;
slow = slow.next;
fast = fast.next.next;
}
//截断左节点
pre.next = null;
//第一次循环后，找到树的root节点位置
TreeNode n = new TreeNode(slow.val);
n.left = sortedListToBST(head); //传递上半区的链表，挂载到左节点上
n.right = sortedListToBST(slow.next); //开始挂右节点
return n;
}
