【LeetCode】108. Convert Sorted Array to Binary Search Tree 解題報告

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108. Convert Sorted Array to Binary Search Tree / Easy

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

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 <= 10^4
  • -10^4 <= nums[i] <= 10^4
  • nums is sorted in a strictly increasing order.

Solution: Recursive

思路

因為題目說要轉換成一顆 height-balanced 的 BST，所以我們只要每次找出中間的 node，再將左邊和右邊的數字分別遞迴即可。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(logN)

LeetCode Result

Code

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class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return buildTree(nums, 0, nums.size()-1);
    }
        
    TreeNode* buildTree(vector<int>& nums, int start, int end) {
        if(end < start) return nullptr;
        int mid = (start+end)/2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = buildTree(nums, start, mid-1); 
        root->right = buildTree(nums, mid+1, end);
        return root;
    }
};