feat(0784_insert-into-a-binary-search-tree): add py3 soln
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0784_insert-into-a-binary-search-tree/README.md
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0784_insert-into-a-binary-search-tree/README.md
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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.
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**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**.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg)
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Input: root = [4,2,7,1,3], val = 5
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Output: [4,2,7,1,3,5]
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Explanation: Another accepted tree is:
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**Example 2:**
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Input: root = [40,20,60,10,30,50,70], val = 25
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Output: [40,20,60,10,30,50,70,null,null,25]
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**Example 3:**
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Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
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Output: [4,2,7,1,3,5]
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**Constraints:**
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* The number of nodes in the tree will be in the range `[0, 104]`.
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* `-108 <= Node.val <= 108`
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* All the values `Node.val` are **unique**.
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* `-108 <= val <= 108`
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* It's **guaranteed** that `val` does not exist in the original BST.
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0784_insert-into-a-binary-search-tree/python3/solution.py
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0784_insert-into-a-binary-search-tree/python3/solution.py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
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if root == None:
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return TreeNode(val)
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current = root
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parent = None
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while True:
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if val < current.val:
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if current.left == None:
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current.left = TreeNode(val)
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break
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parent = current
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current = current.left
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elif val >= current.val:
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if current.right == None:
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current.right = TreeNode(val)
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break
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parent = current
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current = current.right
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return root
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