[LeetCode] 108. Convert Sorted Array to Binary Search Tree ☆(升序数组转换成一个平衡二叉树)

108. Convert Sorted Array to Binary Search Tree

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 every node never differ by more than 1.

题目只有一句话：把一个按升序排列的数组，装换成一个平衡二叉树。

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

解析

对于没有排序的数组来说，就比较麻烦。

但对于已排序的数组，大都用二分递归转换方法来处理。

算法逻辑很简单，就是先找到中间元素，创建根节点，左右子树分别用中间元素左边（即小于中间节点的元素）和中间元素右边（即大于中间节点的元素）递归创建。

这样，节点的左子树永远比节点小，右子树永远比节点大，且由于平均递归创建每一层的子树，所以两个子树的高度差不会超过1。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (null == nums || nums.length == 0) {
            return null;
        }
        return sortedArrayToBSTHelper(nums, 0, nums.length - 1);
    }

    public TreeNode sortedArrayToBSTHelper(int[] nums, int left, int right) {
        if (left > right) {
            return null;
        }
        int mid = (left + right) >> 1;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = sortedArrayToBSTHelper(nums, left, mid - 1);
        node.right = sortedArrayToBSTHelper(nums, mid + 1, right);
        return node;
    }
}