108. Convert Sorted Array to Binary Search Tree

Hint - DFS

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val; // Value of the node
 *     TreeNode *left; // Pointer to the left child
 *     TreeNode *right; // Pointer to the right child
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {} // Default constructor
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} // Constructor with value
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} // Constructor with value and children
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums)
    {
        // Start the recursion with the full range of the array
    }
    
    // Helper function to recursively build the BST
    TreeNode* dfs(vector<int>& nums, int left, int right)
    {
        // Base case: if the current range is invalid, return nullptr
        
        // Find the middle index of the current range
        
        // Create a new node with the value at the middle index
        
        // Recursively build the left subtree with the left half of the range
        
        // Recursively build the right subtree with the right half of the range
        
        // Return the root of the constructed BST
    }
};

Solution - DFS

/**
 * 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)
    {
        return dfs(nums, 0, nums.size() - 1);
    }
    
    TreeNode* dfs(vector<int>& nums, int left, int right)
    {
        if (left > right) return nullptr;
        int mid = (left + right) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = dfs(nums, left, mid - 1);
        root->right = dfs(nums, mid + 1, right);
        return root;
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(\log n)$