Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
分析:
看见二叉树,先想递归。
排序的链表,要转换成平衡二叉树,则中间节点应为root,再递归对前后两段建平衡二叉树就可以了。
找中间节点,则用快慢指针法。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; next = null; }
* }
*/
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
return sortedListToBST(head, null);
}
private TreeNode sortedListToBST(ListNode head, ListNode tail){
if(head==null || head==tail)
return null;
ListNode fast = head;
ListNode slow = head;
while(fast != tail && fast.next != tail){
fast = fast.next.next;
slow = slow.next;
}
TreeNode root = new TreeNode(slow.val);
root.left = sortedListToBST(head, slow);
root.right = sortedListToBST(slow.next, tail);
return root;
}
}