Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.
Example 1:
Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
Example 2:
Input: root = [5,1,7]
Output: [1,null,5,null,7]
Constraints:
- The number of nodes in the given tree will be in the range
[1, 100]. -
0 <= Node.val <= 1000
SOLUTION:
# 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 insert(self, root, key):
if root:
if key <= root.val:
root.left = self.insert(root.left, key)
else:
root.right = self.insert(root.right, key)
else:
root = TreeNode(val = key)
return root
def insort(self, root):
if root:
self.insort(root.left)
self.newroot = self.insert(self.newroot, root.val)
self.insort(root.right)
def increasingBST(self, root: TreeNode) -> TreeNode:
self.newroot = None
self.insort(root)
return self.newroot
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