LeetCode 108. Convert Sorted Array to Binary Search Tree-优快云博客

本文介绍了一种将升序数组转换为高度平衡二叉搜索树的方法。平衡二叉搜索树是一种特殊的二叉树，其中每个节点的两个子树的深度差不超过1。通过递归地选择数组中间元素作为根节点，可以确保树的高度平衡。

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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

     /**
     * Definition for a binary tree node.
     * public class TreeNode {
     * int val;
     * TreeNode left;
     * TreeNode right;
     * TreeNode(int x) { val = x; }
     * }
     */
    public TreeNode sortedArrayToBST(int[] nums) {
        /**
         *根据有序数组构建一颗平衡二叉搜索树
         *就是按照二分查找的顺序构建一颗二叉树
         */
        int low = 0;
        int high = nums.length;
        //递归边界low>high,返回空表明当前节点的子节点为空
        if (low >= high) {
            return null;
        }
        int mid = (low + high) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        //比当前节点小的值作为左孩子，这里没有用mid-1是因为在截取子数组的时候，Arrars.copyOfRange不包含右端点
        node.left = sortedArrayToBST(Arrays.copyOfRange(nums, low, mid));
        //比当前节点值大的作文右孩子
        node.right = sortedArrayToBST(Arrays.copyOfRange(nums, mid + 1, high));
        return node;
    }