【LeetCode】Convert Sorted List to Binary Search Tree 解题报告

标签：平衡二叉树   快慢指针   分治   sorted list   

【题目】

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

【解析】

分治：用快慢指针找到链表的中点，作为树的root，然后二分——中点之前的链表和中点之后的链表分别再构造二叉平衡树。

/**
 * 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;
        if (head.next == null) return new TreeNode(head.val);
        if (head.next.next == null) {
            TreeNode root = new TreeNode(head.val);
            root.right = new TreeNode(head.next.val);
            return root;
        }
        ListNode fast = head;
        ListNode slow = head;
        ListNode preSlow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            preSlow = slow;
            slow = slow.next;
        }
        ListNode mid = slow;
        preSlow.next = null; //make the first part end!!!
        TreeNode root = new TreeNode(mid.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(mid.next);
        return root;
    }
}

容易出错的地方是，中点之前的链表结尾要设为null，表示该链表的结束。

【LeetCode】Convert Sorted List to Binary Search Tree 解题报告

标签：平衡二叉树   快慢指针   分治   sorted list   

原文地址：http://blog.csdn.net/ljiabin/article/details/41451671