701. Insert into a Binary Search Tree - Explanation

Description

You are given the root node of a binary search tree (BST) and a value val to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Note: There may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Example 1:

Input: root = [5,3,9,1,4], val = 6

Output: [5,3,9,1,4,6]

Example 2:

Input: root = [5,3,6,null,4,null,10,null,null,7], val = 9

Output: [5,3,6,null,4,null,10,null,null,7,null,null,9]

Constraints:

0 <= The number of nodes in the tree <= 10,000.
-100,000,000 <= val, Node.val <= 100,000,000
All the values Node.val are unique.
It's guaranteed that val does not exist in the original BST.

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1. Recursion

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if not root:
            return TreeNode(val)

        if val > root.val:
            root.right = self.insertIntoBST(root.right, val)
        else:
            root.left = self.insertIntoBST(root.left, val)

        return root

Time & Space Complexity

Time complexity: $O(h)$
Space complexity: $O(h)$ for the recursion stack.

Where $h$ is the height of the given binary search tree.

2. Iteration

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if not root:
            return TreeNode(val)

        cur = root
        while True:
            if val > cur.val:
                if not cur.right:
                    cur.right = TreeNode(val)
                    return root
                cur = cur.right
            else:
                if not cur.left:
                    cur.left = TreeNode(val)
                    return root
                cur = cur.left

Time & Space Complexity

Time complexity: $O(h)$
Space complexity: $O(1)$ extra space.

Where $h$ is the height of the given binary search tree.