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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
}
};
  • 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;
}
};
  • T: O(n)O(n)
  • S: O(logn)O(\log n)