Convert Sorted Array to Binary Search Tree

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.

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

---

 

Appraoch #1: C++.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if (nums.size() == 0) return nullptr;
        if (nums.size() == 1) return new TreeNode(nums[0]);
        
        int mid = nums.size() / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        
        vector<int> leftNums(nums.begin(), nums.begin()+mid);
        vector<int> rightNums(nums.begin()+mid+1, nums.end());
        
        root->left = sortedArrayToBST(leftNums);
        root->right = sortedArrayToBST(rightNums);
        
        return root;
    }
};

　　

Approach #2: Java.

/**
 * 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 (nums.length < 1)
            return null;
        TreeNode head = helper(nums, 0, nums.length-1);
        return head;
    }
    
    private TreeNode helper(int[] nums, int low, int height) {
        if (low > height) return null;
        int mid = (low + height) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(nums, low, mid-1);
        node.right = helper(nums, mid+1, height);
        return node;
    }
        
}

　　

Approach #3: Python.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if len(nums) == 0:
            return None
        
        mid = len(nums) / 2
        root = TreeNode(nums[mid])
        
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid+1:])
        
        return root