Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-08-18 08:24:28 发布

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本文介绍了一种将有序单链表转换为高度平衡的二叉搜索树的方法。利用快慢指针找到链表中点作为根节点，并递归构造左右子树。

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

思路： Using slow and fast pointer to find the middle pointer of the link list. Using the slow as the root, fast = slow .next; pre.next = null; slow.next = null;

Then recursion to get the left tree and right tree

易错点：注意 slow, pre , fast 全指向head 的情况，要将 head 设置成null 来获得一个 null Treenode 一定要断开 slow 前面和后面的指针。顺序问题，一定要先把 fast = slow.next; 再pre.next = null; 这个关系到全指向head 的情况

/**
 * 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) {
        if(head == null)
            return null;
        ListNode slow = head;
        ListNode fast = head;
        ListNode pre = head;
        while(fast.next != null && fast.next.next != null){
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        TreeNode root = new TreeNode(slow.val);
        fast = slow.next;
        pre.next = null;//-------
        slow.next = null;//-----------
        if(slow == head){//------
            head = null;
        }
        TreeNode left = sortedListToBST(head);
        TreeNode right = sortedListToBST(fast);
        root.left = left;
        root.right = right;
        return root;
    }
}