【leetcode】Convert Sorted Array to Binary Search Tree (easy)

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

有序数组变二叉平衡搜索树，不难，递归就行。每次先序建立根节点（取最中间的数），然后用子区间划分左右子树。

一次就AC了

注意：new 结构体的时候对于

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

要用 TreeNode * root = new TreeNode(123);  与构造函数的形式要一致。

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

//Definition for binary tree
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) {
        if(num.empty())
        {
            return NULL;
        }

        int nr = num.size()/2;
        TreeNode * root = new TreeNode(num[nr]);

        vector<int> numl(num.begin(), num.begin() + nr);
        vector<int> numr(num.begin() + nr + 1, num.end());
        root->left = sortedArrayToBST(numl);
        root->right = sortedArrayToBST(numr);

        return root;
    }
};

int main()
{
    Solution s;
    vector<int> num;
    num.push_back(1);
    num.push_back(2);
    num.push_back(3);
    num.push_back(4);
    num.push_back(5);
    num.push_back(6);
    num.push_back(7);

    TreeNode * ans = s.sortedArrayToBST(num);

    return 0;
}