Convert Sorted Array to Binary Search Tree(将有序数组转为二叉搜索树)

项目github地址：bitcarmanlee easy-algorithm-interview-and-practice
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1.问题描述

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 everynode never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

   0/ \-3   9/   /-10  5

2.解法

二叉搜索树如果按中序遍历，得到的是一个有序数组。那么反过来易知，根节点应该是有序数组的中间点，从中间点分开为左右两个有序数组，在分别找出其中间点作为原中间点的左右两个子节点，本质上是二分查找的思想。

3.java版本

import java.util.*;/*** Created by wanglei on 19/4/14.*/
public class ArrayToBST {public static TreeNode buildTree(int[] nums) {TreeNode root = null;if (nums.length > 0) {int midPos = nums.length / 2;int mid = nums[midPos];root = new TreeNode(mid);root.left = buildTree(Arrays.copyOfRange(nums, 0, midPos));root.right = buildTree(Arrays.copyOfRange(nums, midPos+1, nums.length));}return root;}public static void preTraverse(TreeNode root) {if (root != null) {preTraverse(root.left);System.out.println(root.data);preTraverse(root.right);}}public static void main(String[] args) {int[] nums = {-10, -3, 0, 5, 9};TreeNode root = buildTree(nums);preTraverse(root);}
}

4.python版本

class TreeNode(object):def __init__(self, val):self.val = valself.left = Noneself.right = Nonedef sortedArrayToBST(nums):if nums:midPos = len(nums) / 2mid = nums[midPos]root = TreeNode(mid)root.left = sortedArrayToBST(nums[:midPos])root.right = sortedArrayToBST(nums[midPos+1:])return rootdef printTree(root):if root:printTree(root.left)print root.valprintTree(root.right)nums = [-10, -3, 0, 5, 9]
root = sortedArrayToBST(nums)
printTree(root)