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Insert_Into_A_Binary_Search_Tree.py
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Insert_Into_A_Binary_Search_Tree.py
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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:
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]
# 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
# Iterative Solution
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
if root is None:
return TreeNode(val)
current = root
while True:
if val <= current.val:
if current.left:
current = current.left
else:
current.left = TreeNode(val)
break
else:
if current.right:
current = current.right
else:
current.right = TreeNode(val)
break
return root
# Recursive Solution
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
if root is None:
return TreeNode(val)
elif val <= root.val:
root.left = self.insertIntoBST(root.left, val)
else:
root.right = self.insertIntoBST(root.right, val)
return root