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文章作者:Tyan
博客: | |Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
这个题主要是根据一个有序数组构造二叉查找树(树的左结点小于根节点,根节点小于右结点,子树具有同样的性质)。构造方法主要是递归,每次构建子树时都需要将数组分成左右两半,左边的构建左子树,右边的构建右子树,中间元素构造根节点。
/** * 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 sortedArrayToBST(int[] nums) { return buildBinarySearchTree(nums, 0, nums.length - 1); } public TreeNode buildBinarySearchTree(int[] nums, int start, int end) { if(start > end) { return null; } int mid = (start + end) / 2; TreeNode root = new TreeNode(nums[mid]); root.left = buildBinarySearchTree(nums, start, mid - 1); root.right = buildBinarySearchTree(nums, mid + 1, end); return root; }}
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