LeetCode #108 - Convert Sorted Array to Binary Search Tree-优快云博客

本文介绍了一种将有序数组转换为高度平衡二叉搜索树(BST)的方法。通过递归方式选择中间元素作为根节点，并递归构造左右子树，确保了树的高度平衡。示例代码展示了具体的实现细节。

题目描述：

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

将有序数组构造成BST，采用递归的方法，递归函数需要确定数组下标范围方便调用，然后以下标范围的中点作为根节点。

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