LeetCode/114_Flatten_Binary_Tree_to_Linked_List.py at main · AnmolPatel20/LeetCode · GitHub

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"""
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)

"""

# 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: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        def dfs(root):
            if not root:
                return None

            leftTail = dfs(root.left)
            rightTail = dfs(root.right)

            if root.left:
                leftTail.right = root.right
                root.right = root.left
                root.left = None
            last = rightTail or leftTail or root
            return last
        dfs(root)