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 in python:
# 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 increasingBST(self, root: TreeNode) -> TreeNode:
def traverse(root):
if root == None:
return []
else:
result = []
left = traverse(root.left)
result.extend(left)
result.append(root.val)
right = traverse(root.right)
result.extend(right)
return result
alist = traverse(root)
first = TreeNode(alist[0], None, None)
p = first
for item in alist[1:]:
temp = TreeNode(item, None, None)
p.right = temp
p = temp
return first
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