Leetcode|Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
注意是高度平衡的数。所以选好每个点很关键。不用重新比较。
解法1：利用函数重载实现递归(20ms)

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-1);
        root->right=sortedArrayToBST(nums,mid+1,end);
        return root;
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int n=nums.size();
        TreeNode *root=NULL;
        if(n==0) return root;
        return sortedArrayToBST(nums,0,n-1);
    }

解法2：直接利用迭代器来递归(24ms)

TreeNode* sortedArrayToBST(vector<int>& nums){//用一个函数实现，不重载
        int n=nums.size();
        TreeNode *root=NULL;
        if(n==0) return root;
        int mid=n/2;
        root=new TreeNode(nums[mid]);
        vector<int>::iterator first,last;
        first=nums.begin();
        last=nums.begin()+mid;
        vector<int> v(first,last);
        root->left=sortedArrayToBST(v);
        if(mid<n){
           first=nums.begin()+mid+1;
           last=nums.end();
           vector<int> w(first,last);
           root->right=sortedArrayToBST(w);
        }
        return root;
    }