LeetCode 897. Increasing Order Search Tree

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)

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