108. Convert Sorted Array to Binary Search Tree

题目：

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

链接： http://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

题解：

排序好的数组转换为BST, 找中点即可。

Time Complexity - O(n)， Space Complexity - O(n)。

/**
 * 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) {
        if(nums == null || nums.length == 0)
            return null;
        return sortedArrayToBST(nums, 0, nums.length - 1);
    }
    
    private TreeNode sortedArrayToBST(int[] nums, int lo, int hi) {
        if(lo > hi)
            return null;
        int mid = lo + (hi - lo) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = sortedArrayToBST(nums, lo, mid - 1);
        root.right = sortedArrayToBST(nums, mid +1, hi);
        return root;
    } 
}

 

二刷:

一刷代码比较简洁，不像自己写的...

主要我们就是利用一个辅助方法，使用二分法来自顶向下递归构建树。也可以使用自底向上，另外一题Convert Sorted List to BST就适合用自底向上方法。

Java:

Time Complexity - O(n)， Space Complexity - O(n)。

/**
 * 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) {
        if (nums == null || nums.length == 0) return null;
        return buildTree(nums, 0, nums.length - 1);
    }
    
    private TreeNode buildTree(int[] nums, int lo, int hi) {
        if (lo > hi) return null;
        int mid = lo + (hi - lo) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = buildTree(nums, lo, mid - 1);
        node.right = buildTree(nums, mid + 1, hi);
        return node;
    }
}

 

 

测试：