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二叉搜索树(BST)的方法。核心思路在于利用中点分割的方式，选取链表中间元素作为BST根节点，并递归地将左右两部分链表分别构建为根节点的左右子树。

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

思路：找到中间节点，中间节点是BST树根，然后中间节点的左半部分是树根的左子树，右半部分是树根的右子树，依次递归下去。

/**
 * 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; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) 
    {
        int nlen = 0,i=0;
        ListNode p =head;
        while(p!=null)
        {
            nlen++;
            p=p.next;
        }
        if(nlen==0)
            return null;
        int[] num = new int[nlen];
        while(head!=null)
        {
            num[i++]=head.val;
            head=head.next;
        }
        int mid = nlen/2;
        TreeNode root = new TreeNode(num[mid]);
        helper(num,0,mid-1,root,0);
        helper(num,mid+1,nlen-1,root,1);
        return root;
    }
    public void helper(int[] num,int i,int j,TreeNode cur,int flag)
    {
        if(i<=j)
        {
            int l = i, r = j;
            int mid = l+(r-l)/2;
            TreeNode tn = new TreeNode(num[mid]);
            if(flag==0) // left
            {
                cur.left=tn;
                helper(num,l,mid-1,tn,0);
                helper(num,mid+1,r,tn,1);
            }
            if(flag==1) // right
            {
                cur.right=tn;
                helper(num,l,mid-1,tn,0);
                helper(num,mid+1,r,tn,1);
            }
        }
    }
}