Convert Sorted Array to Binary Search Tree 把一个有序数组转换成BST @LeetCode

题目：

给定一个有序的数组，要求转换为BST

思路：

递归。

每次找到数组中值作为BST的root，然后递归处理左半数组和右半数组分别作为左右子树

package Level2;import Utility.TreeNode;/** * Convert Sorted Array to Binary Search Tree   *  * Given an array where elements are sorted in ascending order, convert it to a height balanced BST. */public class S108 {public static void main(String[] args) {int[] num = {1,3};TreeNode n = sortedArrayToBST(num);n.print();}public static TreeNode sortedArrayToBST(int[] num) {return sortedArrayToBSTRec(num, 0, num.length-1);    }// 二分查找节点值，并递归创建节点private static TreeNode sortedArrayToBSTRec(int[] num, int low, int high){if(low > high){return null;}int mid = low + (high-low)/2;// 找到中值TreeNode root = new TreeNode(num[mid]);// 创建根节点        root.left = sortedArrayToBSTRec(num, low, mid-1);// 递归创建左子树和右子树        root.right = sortedArrayToBSTRec(num, mid+1, high);return root;}}

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode sortedArrayToBST(int[] num) {        return rec(num, 0, num.length-1);    }        public TreeNode rec(int[] num, int low, int high){        if(low > high){            return null;        }        int mid = low + (high-low)/2;        TreeNode root = new TreeNode(num[mid]);        root.left = rec(num, low, mid-1);        root.right = rec(num, mid+1, high);        return root;    }}