Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-08-18 08:24:28 发布

最新推荐文章于 2020-08-18 08:24:28 发布 · 56 阅读

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本文详细介绍了如何将一个元素按升序排列的单链表转换为一个高度平衡的二叉搜索树。通过递归地选择中间节点作为根节点，其余节点递归地构建左右子树，确保了树的高度平衡。

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

/**
 * 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 {
	private ListNode h;
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
        	return null;
        }
        h = head;
        ListNode tmp = head;
        int cnt = 0;
        while (tmp != null) {
        	cnt++;
        	tmp = tmp.next;
        }
        return solve(0, cnt-1);
    }
	private TreeNode solve(int s, int e) {
		if (s > e) {
			return null;
		}
		int mid = (s+e)/2;
		TreeNode left = solve(s, mid-1);
		TreeNode root = new TreeNode(h.val);
		root.left = left;
		h = h.next;
		TreeNode right = solve(mid+1, e);
		root.right = right;
		return root;
	}
}