LeetCode-in-All

114. Flatten Binary Tree to Linked List

Medium

Given the root of a binary tree, flatten the tree into a “linked list”:

• The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]

Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []

Output: []

Example 3:

Input: root = [0]

Output: [0]

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

To solve the “Flatten Binary Tree to Linked List” problem in Python with a Solution class, we’ll use a recursive approach. Below are the steps:

1. Create a Solution class: Define a class named Solution to encapsulate our solution methods.

2. Create a flatten method: This method takes the root node of the binary tree as input and flattens the tree into a linked list using preorder traversal.

3. Check for null root: Check if the root is null. If so, there’s no tree to flatten, so return.

4. Recursively flatten the tree: Define a recursive helper method flattenTree to perform the flattening.
   • The method should take the current node as input.
   • Perform a preorder traversal of the tree.
   • For each node, if it has a left child:
     • Find the rightmost node in the left subtree.
     • Attach the right subtree of the current node to the right of the rightmost node.
     • Move the left subtree to the right subtree position.
     • Set the left child of the current node to null.
   • Recursively call the method for the right child.
5. Call the helper method: Call the flattenTree method with the root node.

Here’s the Python implementation:

# 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 flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if root:
            self.find_tail(root)

    def find_tail(self, root: TreeNode) -> TreeNode:
        left = root.left
        right = root.right
        # Find the tail of the left subtree, tail means the most left leaf
        if left:
            tail = self.find_tail(left)
            # Stitch the right subtree below the tail
            root.left = None
            root.right = left
            tail.right = right
        else:
            tail = root
        # Find tail of the right subtree
        if tail.right is None:
            return tail
        else:
            return self.find_tail(tail.right)   

This implementation follows the steps outlined above and efficiently flattens the binary tree into a linked list using preorder traversal in Python.

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