## Convert Sorted Array to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

### Analysis:

Because the requirement "height balanced", this problem becomes relative easy.
Just recursively use the middle element in the array as the root node of the current tree(sub tree).

### Code(C++):

/**
* 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* cbst(vector<int> &num, int st, int ed){
if (st>ed){
return NULL;
}else{
int mid = st+(ed-st)/2;
TreeNode *bst=new TreeNode(num[mid]);
bst->left = cbst(num,st,mid-1);
bst->right = cbst(num,mid+1,ed);
return bst;
}
}

TreeNode *sortedArrayToBST(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (num.size()==0){return NULL;}
return cbst(num,0,num.size()-1);
}
};


### Code(Python):

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
# @param num, a list of integers
# @return a tree node
def bst(self, num, st, ed):
if st > ed:
return None
mid = st + (ed - st)/2
left = self.bst(num, st, mid - 1)
root = TreeNode(num[mid])
right = self.bst(num, mid + 1, ed)
root.left = left
root.right = right
return root

def sortedArrayToBST(self, num):
n = len(num)
if n == 0:
return None
return self.bst(num, 0, n-1)


#### 1 comment:

1. You usually have Java code for the algorithms, solution is trivial still thought I'd comment in mine

public class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
if(nums.length==0) return null;
return createBST(0,nums.length-1,nums);
}

private TreeNode createBST(int start, int end, int[] nums) {
if(start>end) return null;
else {
int mid=start+ end >>1;
TreeNode subTree=new TreeNode(nums[mid]);
subTree.left=createBST(start,mid-1,nums);
subTree.right=createBST(mid+1,end,nums);
return subTree;
}
}
}

if the question asked to create a complete binary tree i.e. all leaf nodes must be as far left as possible could you help as to how to approach that?