108 Convert Sorted Array to Binary Search Tree

It’s crucial that while offloading nodes to the list, we do an inorder() traversal - as that would visit first the left side nodes, then the root and finally the right side nodes.

1382 Balance a Binary Search Tree uses the fundamental knowledge that you build in this problem. I recommend solving that too while you are on this one.

Code

Python3

def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
 
    def construct_tree(start: int, end: int) -> Optional[TreeNode]:
        if start > end: return None
 
        mid = (start + end) // 2
 
        cur_root = TreeNode(nums[mid])
 
        cur_root.left = construct_tree(start, mid - 1)
        cur_root.right = construct_tree(mid + 1, end)
 
        return cur_root
 
    return construct_tree(0, len(nums)-1)

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
    return constructBST(0, len(nums)-1, nums)
}
 
func constructBST(left, right int, arr []int) *TreeNode {
    if left > right {
        return nil
    }
 
    mid := (left + right) / 2
    root := &TreeNode{Val: arr[mid]}
 
    root.Left = constructBST(left, mid - 1, arr)
    root.Right = constructBST(mid + 1, right, arr)
 
    return root
}

Big O Analysis

• Runtime

  The runtime complexity here is $O(N)$ since we would have to visit all nodes in the array atleast once to construct the tree.

• Memory

  The memory usage is $O(N)$ since we are creating a new TreeNode for every value in the array.

— A

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