108. Convert Sorted Array to Binary Search Tree

1.問題

• 利用給予的序列建構出balanced binary search tree

2.想法

• 提問

  • 確認題意: height balanced

• function header, parameter input

• 說明想法

  • BST定義為left descents <= n < right descents, 如果用二分法, 將每次的mid作為新的root node, 不僅左右高度相同, 而且滿足BST的定義

• 測試計算複雜度

3.程式碼

/**
 * 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 NULL;
        }
        return buildTree(nums, 0, nums.size() - 1);
    }
private:
    TreeNode* buildTree(vector<int>& nums, int left, int right) {
        if (left > right) {
            return NULL;
        }
        
        int mid = (left + right) / 2;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = buildTree(nums, left, mid - 1);
        node->right = buildTree(nums, mid + 1, right);
        
        return node;
    }
};