leetcode

题目：Convert Sorted Array to Binary Search Tree

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

个人思路：

1、选取数组的中点作为根节点，中点左半边的数作为根节点的左子树节点，右半边的数作为根节点的右子树节点，递归地对左子树和右子树进行处理

2、递归结束条件为数组的起始点大于结束点

代码：

#include <stddef.h>
#include <vector>
#include <iostream>

using namespace std;

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    TreeNode *sortedArrayToBST(vector<int> &num)
    {
        TreeNode *root = build(num, 0, num.size() - 1);
        //cout << root->val << endl;
        //preorder(root);
        return root;
    }
    void preorder(TreeNode *root)
    {
        if (root == NULL)
        {
            return;
        }
        cout << root->val << endl;
        preorder(root->left);
        preorder(root->right);
    }
private:
    TreeNode *build(vector<int> &num, int start, int end)
    {
        if (start > end)
        {
            return NULL;
        }

        int middle = (start + end) / 2;
        TreeNode *treeNode = new TreeNode(num[middle]);
        treeNode->left = build(num, start, middle - 1);
        treeNode->right = build(num, middle + 1, end);

        //cout << treeNode->val << endl;

        return treeNode;
    }
};

int main()
{
    Solution s;
    vector<int> v;
    v.push_back(1);
    v.push_back(2);
    v.push_back(3);
    v.push_back(4);
    v.push_back(5);
    v.push_back(6);

    TreeNode *root = s.sortedArrayToBST(v);
    s.preorder(root);

    system("pause");
    return 0;
}

网上的方法都是类似的