Understanding the Problem
We are given a Binary Tree, where nodes can have values in any order. Unlike a Binary Search Tree (BST), a Binary Tree doesn't guarantee that all left descendants are less than the root and all right descendants are greater. Our goal is to convert this Binary Tree into a BST, but without changing its structure.
This means that we can only change the node values, not the arrangement of nodes. The final tree should look the same in shape but must satisfy the BST property with the new values.
Step-by-Step Solution with Example
Step 1: Perform Inorder Traversal to Collect Values
Inorder traversal visits nodes in the left-root-right order. For any binary tree, this traversal doesn't guarantee sorted output, but it will help us collect all node values in a list.
Step 2: Sort the Collected Values
Once we have all the values in a list, we sort the list in ascending order. This sorted list represents how values should appear in inorder traversal for a valid BST.
Step 3: Perform Inorder Traversal Again to Replace Node Values
This time, during the inorder traversal, instead of collecting values, we replace each node’s value with the next value from the sorted list. Since inorder traversal visits nodes in the correct BST order (left-root-right), replacing values in this sequence ensures the BST property is satisfied.
Step 4: Example
Let's say we have the following binary tree:
10
/ 30 15
/ 20 5
Inorder traversal of this tree: [20, 30, 10, 15, 5]
Sort the list: [5, 10, 15, 20, 30]
Replace values during inorder traversal:
20
/ 10 30
/ 5 15
The structure remains the same, but now the values follow the BST rules.
Edge Cases
Case 1: Tree Already a BST
If the tree is already a BST, its inorder traversal is already sorted. Sorting and replacing with the same values has no effect. The operation is idempotent.
Case 2: Single Node Tree
A tree with just one node is trivially a BST. The algorithm still works and leaves the node unchanged.
Case 3: Empty Tree
An empty tree has no nodes to convert. The algorithm should detect this and return immediately. This ensures robustness.
Case 4: Unbalanced Tree
In an unbalanced tree (e.g., all nodes skewed to one side), the traversal and value replacement still work because the logic depends only on traversal, not shape.
Finally
This method of converting a binary tree to a BST is elegant and beginner-friendly. It works by rearranging node values rather than node positions, making it less complex. It also handles all edge cases gracefully, making the solution robust and widely applicable in real-world scenarios.
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