Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
从给定的有序链表生成平衡二叉树。
解题思路：最容易想到的就是利用数组生成二叉树的方法，找到中间节点作为二叉树的root节点，然后分别对左右链表递归调用分别生成左子树和右子树。时间复杂度O(N*lgN)
AC代码：
public class Solution {
ListNode getLeftNodeFromList(ListNode head) {
ListNode next = head;
ListNode current = head;
ListNode pre = head;
while(next!=null) {
next = next.next;
if(next==null) {
break;
}
next = next.next;
if(next==null) {
break;
}
pre = head;
head = head.next;
}
return pre;
}
public TreeNode sortedListToBST(ListNode head) {
if(head==null) {
return null;
}
if(head.next==null) {
return new TreeNode(head.val);
}
ListNode left = getLeftNodeFromList(head);
ListNode mid = left.next;
TreeNode root = new TreeNode(mid.val);
left.next = null;
root.left = sortedListToBST(head);
root.right = sortedListToBST(mid.next);
return root;
}
}
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