LeetCode—109. Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-01-12 00:53:34 发布

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本文介绍如何将一个递增的链表转换为平衡二叉搜索树，提供了两种方法：一种是先将链表转化为数组，再利用数组转换为平衡二叉树；另一种是在链表上直接进行转换。通过递归方式创建平衡二叉树，确保了树的高度尽可能小。

LeetCode—109. Convert Sorted List to Binary Search Tree

题目

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/

给出一个递增的链表，将其转化成一个平衡二叉搜索树。
在这里插入图片描述

思路及解法

与数组转化成平衡二叉树相似。可以直接将list转化成数组，然后用数组转化的方法。当然也可以直接转化。

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head==null) return null;
        int len  = 1;
        ListNode listNode  = head;
        while(listNode.next != null){
            listNode = listNode.next;
            len++;
        }
        listNode = head;
        int[] nums = new int[len];
        for(int i=0; i<len; i++){
            nums[i] = listNode.val;
            listNode = listNode.next;
        }
        TreeNode node = createTree(nums, 0, len-1);
        return node;
    }
    public TreeNode createTree(int[] nums, int s, int e){
        if(s>e) return null;
        int mid = (s+e)/2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = createTree(nums, s, mid-1);
        node.right = createTree(nums, mid+1, e);
        return node;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head==null) return null;
        int len  = 1;
        ListNode listNode = head;
        while(listNode.next != null){
            len++;
            listNode = listNode.next;
        }
        TreeNode node = createTree(head, 0, len-1);
        return node;
    }
    public TreeNode createTree(ListNode ln, int s, int e){
        if(s>e) return null;
        int mid = (s+e)/2;
        ListNode midNode = ln;
        for(int i=0; i<mid; i++){
            midNode = midNode.next;
        }
        TreeNode node = new TreeNode(midNode.val);
        node.left = createTree(ln, s, mid-1);
        node.right = createTree(ln, mid+1, e);
        return node;
    }
}