feat(0784_insert-into-a-binary-search-tree): add py3 soln

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Sangeeth Sudheer 2022-04-05 22:02:10 +05:30
parent 65c530be33
commit b32128e01c
2 changed files with 64 additions and 0 deletions

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You are given the `root` node of a binary search tree (BST) and a `value` 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.
**Notice** that 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:**
![](https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg)
Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:
**Example 2:**
Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]
**Example 3:**
Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]
**Constraints:**
* The number of nodes in the tree will be in the range `[0, 104]`.
* `-108 <= Node.val <= 108`
* All the values `Node.val` are **unique**.
* `-108 <= val <= 108`
* It's **guaranteed** that `val` does not exist in the original BST.

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# 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 root == None:
return TreeNode(val)
current = root
parent = None
while True:
if val < current.val:
if current.left == None:
current.left = TreeNode(val)
break
parent = current
current = current.left
elif val >= current.val:
if current.right == None:
current.right = TreeNode(val)
break
parent = current
current = current.right
return root