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Convert Sorted List to Binary Search Tree

描述​

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

分析​

这题与上一题类似，但是单链表不能随机访问，而自顶向下的二分法必须需要 RandomAccessIterator，因此前面的方法不适用本题。

存在一种自底向上(bottom-up)的方法，见 http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

分治法，自顶向下​

分治法，类似于 Convert Sorted Array to Binary Search Tree，自顶向下，复杂度 O(nlogn)。

// Convert Sorted List to Binary Search Tree
// 二分法，类似于 Convert Sorted Array to Binary Search Tree，
// 自顶向下，时间复杂度O(nlogn)，空间复杂度O(logn)
class Solution {
    public TreeNode sortedListToBST (ListNode head) {
        if(head == null) return null;
        if(head.next == null) return new TreeNode(head.val);

        ListNode mid = cutAtMiddle(head);

        TreeNode root = new TreeNode(mid.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(mid.next);

        return root;
    }

    ListNode cutAtMiddle(ListNode head) {
        if(head == null) return null;

        ListNode fast = head;
        ListNode slow = head;
        ListNode prev_slow = head;

        while(fast != null && fast.next != null){
            prev_slow = slow;
            slow = slow.next;
            fast = fast.next.next;
        }

        prev_slow.next = null;
        return slow;
    }
}

自底向上​

// Convert Sorted List to Binary Search Tree
// bottom-up，时间复杂度O(n)，空间复杂度O(logn)
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        int len = 0;
        ListNode p = head;
        while (p != null) {
            len++;
            p = p.next;
        }
        return sortedListToBST(new Container(head), 0, len - 1);
    }
    private static TreeNode sortedListToBST(Container list, int start, int end) {
        if (start > end) return null;

        int mid = start + (end - start) / 2;
        TreeNode leftChild = sortedListToBST(list, start, mid - 1);
        TreeNode parent = new TreeNode(list.p.val);
        parent.left = leftChild;
        list.p = list.p.next;
        parent.right = sortedListToBST(list, mid + 1, end);
        return parent;
    }
    // 模拟二级指针
    static class Container {
        ListNode p;
        public Container(ListNode p) {
            this.p = p;
        }
    }
}

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