108. Convert Sorted Array to Binary Search Tree

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

二叉树的问题多半是要使用递归，提供一个完整的可编译运行的程序，可以调试运行，学习运行过程:

#include <iostream>
#include <vector>
using namespace std;

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) {}
 };

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

TreeNode* sortedArrayToBST(vector<int>& nums) 
{
    return sortedArrayToBST(nums,0,nums.size());
}

void printBST(TreeNode * root)
{
    if(NULL == root)
        return;
    cout << " " << root->val <<" ";
    printBST(root->left);
    printBST(root->right);

}
int main()
{
    vector <int> nums{-10,-3,0,5,9};
    TreeNode *res = sortedArrayToBST(nums);
    printBST(res);
    return 0;
}