有序链表转换二叉搜索树

给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

给定的有序链表： [-10, -3, 0, 5, 9],

一个可能的答案是：[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树：

      0
     / \
   -3   9
   /   /
 -10  5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

找中点，递归建树

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {

        if (head == null) return null;
        return buildTree(head, null);

    }


    public TreeNode buildTree(ListNode head, ListNode tail) {

        if (head == tail) {
            return null;
        }
        ListNode mid = getMidNode(head, tail);
        TreeNode root = new TreeNode(mid.val);

        TreeNode left = buildTree(head, mid);
        TreeNode right = buildTree(mid.next, tail);
        root.left = left;
        root.right = right;
        return root;
    }

    public ListNode getMidNode(ListNode head, ListNode tail) {
        ListNode fast = head;
        ListNode slow = head;

        while (fast != tail && fast.next != tail) {
            fast = fast.next;
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

中序遍历优化

class Solution {
    ListNode globalHead;

    public TreeNode sortedListToBST(ListNode head) {
        globalHead = head;
        int length = getLength(head);
        return buildTree(0, length - 1);
    }

    public int getLength(ListNode head) {
        int ret = 0;
        while (head != null) {
            ++ret;
            head = head.next;
        }
        return ret;
    }

    public TreeNode buildTree(int left, int right) {
        if (left > right) {
            return null;
        }
        int mid = (left + right + 1) / 2;
        TreeNode root = new TreeNode();
        root.left = buildTree(left, mid - 1);
        root.val = globalHead.val;
        globalHead = globalHead.next;
        root.right = buildTree(mid + 1, right);
        return root;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree/solution/you-xu-lian-biao-zhuan-huan-er-cha-sou-suo-shu-1-3/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

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