Convert Sorted List to Binary Search Tree 把有序链表转为BST@LeetCode_convert sorted list to binary search tree 将有序链表转为二-优快云博客

本文链接：https://blog.youkuaiyun.com/fightforyourdream/article/details/16940205

本文介绍了一种通过递归方法将排序单链表转换为高度平衡的二叉搜索树的技术。利用快慢指针在链表中寻找中点，从而在每个迭代中构建树的节点。

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万能的递归法，注意到有一点比较巧的是，由于单链表都是单向的。所以这里我的查找区间设置为左闭右开！这样就避免了去找mid节点的前一个节点

package Level4;

import Utility.ListNode;
import Utility.TreeNode;


/**
 * Convert Sorted List to Binary Search Tree
 * 
 *  Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
 *
 */
public class S109 {

	public static void main(String[] args) {
		ListNode head = new ListNode(1);
		ListNode n2 = new ListNode(3);
		head.next = n2;
		TreeNode root = sortedListToBST(head);
	}

	public static TreeNode sortedListToBST(ListNode head) {
        return rec(head, null);
    }
	
	// 在区间[start, end)里递归，后面的end是包括在内的，这样可以避免要多用一个指针来记录mid前的节点
	public static TreeNode rec(ListNode start, ListNode end){
		if(start == end){
			return null;
		}
		
		// 一次遍历找到中点的方法：快慢指针
		ListNode mid = start;			// 该指针最终会指向中点
		ListNode probe = start;			// 探针最终会到达end
		while(probe!=end && probe.next!=end){		// 探针完成搜索，注意停止条件是和end比较而不是和null比！
			mid = mid.next;
			probe = probe.next.next;
		}
		
		TreeNode root = new TreeNode(mid.val);
		root.left = rec(start, mid);
		root.right = rec(mid.next, end);
		return root;
	}

}

/**
 * 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 rec(head, null);
    }
    
    public TreeNode rec(ListNode start, ListNode end) {
        if(start == end) {
            return null;
        }
        ListNode p = start, q = start;
        while(q != end && q.next != end) {
            p = p.next;
            q = q.next.next;
        }
        
        TreeNode root = new TreeNode(p.val);
        root.left = rec(start, p);
        root.right = rec(p.next, end);
        
        return root;
    }
    
}