Is Increasing Order Search Tree easy or hard?

Increasing Order Search Tree is classified as an Easy problem on LeetCode with a high acceptance rate around 78%. The challenge mainly tests understanding of BST in-order traversal and basic tree pointer manipulation.

Increasing Order Search Tree Python/Java solution

Python and Java solutions typically implement recursive in-order traversal with a dummy node pointer to rebuild the tree. Each visited node has its left child set to null and is appended to the right of the previous node, producing the required increasing order tree.

How to solve Increasing Order Search Tree in O(n)?

Perform an in-order traversal of the BST so nodes are processed in sorted order. Maintain a pointer to the previously added node and attach the current node to its right while clearing the left pointer. Since each node is processed once, the total time complexity is O(n).

What is the best approach for Increasing Order Search Tree?

The most common approach uses in-order traversal of the BST and rebuilds the tree during traversal. Because in-order traversal visits nodes in sorted order, you can attach each node to the right of the previous node while setting left pointers to null. This runs in O(n) time and O(h) space where h is the tree height.

Is Increasing Order Search Tree asked at Google/Amazon/Meta?

Problems involving BST traversal and tree restructuring frequently appear in interviews at companies like Amazon, Google, and Meta. This specific problem tests understanding of in-order traversal, pointer manipulation, and binary tree restructuring patterns.

What data structure is used in Increasing Order Search Tree?

The core data structure is a Binary Search Tree. The solution relies on in-order traversal, which may use recursion, an explicit stack, or Morris traversal to visit nodes in sorted order while rebuilding the tree structure.

What is the time complexity of Increasing Order Search Tree?

Both common solutions run in O(n) time because each node in the tree is visited exactly once during the in-order traversal. The recursive version uses O(h) auxiliary space for the recursion stack, while the Morris traversal variant reduces space to O(1).

Description Solution Video 4:32

897. Increasing Order Search Tree

Easy79.1% acceptance2 companies

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

Asked at 2 companies

Google Amazon

Topics

4:32

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1# Definition for a binary tree node.
2# class TreeNode:
3#     def __init__(self, val=0, left=None, right=None):
4#         self.val = val
5#         self.left = left
6#         self.right = right
7class Solution:
8    def increasingBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
9        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[5,3,6,2,4,null,8,1,null,null,null,7,9]

Input

[5,1,7]

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