给出一个所有元素以升序排序的单链表,将它转换成一棵高度平衡的二分查找树
2
1->2->3 => / \
1 3
3
1->2->3->4->5->6 => / \
1 5
/ \ / \
# 2 4 6
本题要求是高度平衡的二叉树,那就看作是标准的平衡二叉树。 首先平衡二叉树要求左右子树的高度差不超过 1,我们把有序列表的中间节点作为根,即可保证左右子树的元素个数相差不超过1,只需要把每一个节点都看作是一棵树,递归取中间节点即可。
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @return: a tree node
*/
public TreeNode sortedListToBST(ListNode head) {
if (head == null) {
return null;
}
//先将链表转化为List,方便获取长度,并随机读取。
List<Integer> list = new ArrayList<Integer>();
while (head != null) {
list.add(head.val);
head = head.next;
}
return toTreeNode(list, 0, list.size()-1);
}
public TreeNode toTreeNode (List<Integer> list, int s, int e) {
//返回条件
if (s > e)
return null;
int mid = (s + e) / 2;
//把中间节点作为根
TreeNode root = new TreeNode(list.get(mid));
//分别递归左右子树
root.left = toTreeNode(list, s, mid - 1);
root.right = toTreeNode(list, mid + 1, e);
return root;
}
}