Description
https://leetcode.com/problems/increasing-order-search-tree/
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
Explanation
Do an in-order traverse binary search tree to find out ascending order values first. Then create a new increasing order tree.
Python 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 increasingBST(self, root: TreeNode) -> TreeNode:
self.orders = []
self.traverse(root)
new_root = TreeNode(self.orders[0])
node = new_root
for val in self.orders[1:]:
node.right = TreeNode(val)
node = node.right
return new_root
def traverse(self, root):
if not root:
return
self.traverse(root.left)
self.orders.append(root.val)
self.traverse(root.right)
- Time Complexity: O(N)
- Space Complexity: O(N)