109. Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-04-13 06:45:19 发布

原创最新推荐文章于 2020-04-13 06:45:19 发布 · 154 阅读

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本文介绍如何将一个按升序排列的单链表转换为高度平衡的二叉搜索树（BST）。提供两种解决方案：一种是先将链表元素转化为数组，再通过数组构建平衡BST；另一种是在O(1)空间复杂度和O(n)时间复杂度下直接处理链表。

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

可以把链表的数取出来变成数组，按照<<Convert Sorted Array to Binary Search Tree>>的办法处理

public class Solution {
    public TreeNode sortedListToBST(ListNode head)
	{
		int[] nums=new int[30000];
		int cnt=0;
		while(head!=null)
		{
			nums[cnt++]=head.val;
			head=head.next;
		}
		return sortedArrayToBST(nums, cnt);
	}

	private TreeNode sortedArrayToBST(int[] nums,int len)
	{
		if (len == 0)
			return null;
		return build(0, len - 1, nums);
	}

	private TreeNode build(int start, int end, int[] nums)
	{
		if (start == end)
			return new TreeNode(nums[start]);
		if (start > end)
			return null;
		int mid = (start + end) >> 1;
		int num = nums[mid];
		TreeNode t = new TreeNode(num);
		t.left = build(start, mid - 1, nums);
		t.right = build(mid + 1, end, nums);
		return t;
	}
}

另外不取出来直接处理的版本

https://discuss.leetcode.com/topic/8141/share-my-o-1-space-and-o-n-time-java-code/2

private ListNode node;

public TreeNode sortedListToBST(ListNode head) {
	if(head == null){
		return null;
	}
	
	int size = 0;
	ListNode runner = head;
	node = head;
	
	while(runner != null){
		runner = runner.next;
		size ++;
	}
	
	return inorderHelper(0, size - 1);
}

public TreeNode inorderHelper(int start, int end){
	if(start > end){
		return null;
	}
	
	int mid = start + (end - start) / 2;
	TreeNode left = inorderHelper(start, mid - 1);
	
	TreeNode treenode = new TreeNode(node.val);
	treenode.left = left;
	node = node.next;

	TreeNode right = inorderHelper(mid + 1, end);
	treenode.right = right;
	
	return treenode;
}