LeetCode 114. Flatten Binary Tree to Linked List



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]


  • 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)?


Recursively find the leftmost leaf node and start from there to add the root.right and replace root.right with root.left.

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 flatten(self, root: TreeNode) -> None:
        Do not return anything, modify root in-place instead.
    def helper(self, root):
        if not root:
            return None
        if not root.left and not root.right:
            return root

        left_tail = self.helper(root.left)
        right_tail = self.helper(root.right)
        if left_tail:            
            left_tail.right = root.right
            root.right = root.left
            root.left = None
        if right_tail:
            return right_tail
        return left_tail
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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