109. 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。

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。

分治（未优化版）

class Solution {

    private ListNode findMid(ListNode head, ListNode end) {
        if (head == end || head.next == end || head.next.next == end) {
            return head;
        }
        ListNode slow = head.next;
        ListNode fast = head.next.next;
        while (fast.next != end && fast.next.next != end) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    private TreeNode solve(ListNode start, ListNode end) {
        /**
         * 空节点
         */
        if (start == end) {
            return null;
        }

        ListNode mid = findMid(start, end);
        TreeNode root = new TreeNode(mid.val);
        root.left = solve(start, mid);
        root.right = solve(mid.next, end);
        return root;
    }

    public TreeNode sortedListToBST(ListNode head) {
        return solve(head, null);
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }

    @Override
    public String toString() {
        return "ListNode{" +
                "val=" + val +
                ", next=" + next +
                '}';
    }
}

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;
    }
}

分治（优化版）

class Solution {

    private ListNode globalNode;

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

    private TreeNode solve(int start, int end) {
        /**
         * 空节点
         */
        if (start > end) {
            return null;
        }

        int mid = (start + end) / 2;
        TreeNode root = new TreeNode();
        root.left = solve(start, mid - 1);
        root.val = globalNode.val;
        globalNode = globalNode.next;
        root.right = solve(mid + 1, end);
        return root;
    }

    public TreeNode sortedListToBST(ListNode head) {
        this.globalNode = head;
        return solve(0, getLength(head) - 1);
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }

    @Override
    public String toString() {
        return "ListNode{" +
                "val=" + val +
                ", next=" + next +
                '}';
    }
}

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;
    }
}

标签：head,ListNode,val,int,next,链表,109,TreeNode,二叉
来源： https://www.cnblogs.com/tianyiya/p/15706613.html