LeetCode 108. Convert Sorted Array to Binary Search Tree （将有序数组转换成BST）

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

/**
 * 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;
        if (nums.size() == 1)
        {
            return new TreeNode(nums[0]);
        }
        
        int mid = nums.size() / 2;
        TreeNode *root = new TreeNode(nums[mid]);
        
        vector<int> leftNs(nums.begin(), nums.begin() + mid);     
        vector<int> rightNs(nums.begin() + mid + 1, nums.end());
        
        root->left = sortedArrayToBST(leftNs);
        root->right = sortedArrayToBST(rightNs);
        
        return root;
    }
};

 法二：

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

};