Convert Sorted Array to Binary Search Tree

Description

Given an integer array nums where the elements are sorted in ascending order, convert it to a

height-balanced

binary search tree.

Example 1:

Input: nums = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

Input: nums = [1,3] Output: [3,1] Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums is sorted in a strictly increasing order.

Code

因為 array 為 sorted，因此中間的 node 就會是 head。

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

Source

• Convert Sorted Array to Binary Search Tree - LeetCode