LeetCode OJ 108. Convert Sorted Array to Binary Search Tree DFS求解

    题目链接:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

108. Convert Sorted Array to Binary Search Tree

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Total Accepted: 68378 Total Submissions: 187560 Difficulty: Medium

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

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    非常有意思的一道题目。

要求依据一个有序数组,构造出一棵高度平衡的BST。

    每次找到数组的中间位置。这个便是BST的 根节点。左右孩子也非常好找,根节点左边区域的中间节点便是左孩子。根节点的右边区域的中间节点便是右孩子。如此递归求解。

    我的AC代码

public class ConvertSortedArraytoBinarySearchTree {

	public TreeNode sortedArrayToBST(int[] nums) {
		return dfs(nums, 0, nums.length - 1);
	}

	TreeNode dfs(int[] nums, int left, int right) {
		if (left > right) return null;
		int mid = (left + right) / 2;
		TreeNode root = new TreeNode(nums[mid]);
		root.left = dfs(nums, left, mid - 1);
		root.right = dfs(nums, mid + 1, right);
		return root;
	}
}


原文地址:https://www.cnblogs.com/wgwyanfs/p/7091371.html