114-flatten-binary-tree-to-linked-list¶
Try it on leetcode
Description¶
Given the root
of a binary tree, flatten the tree into a "linked list":
- The "linked list" should use the same
TreeNode
class where theright
child pointer points to the next node in the list and theleft
child pointer is alwaysnull
. - 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)?Solution(Python)¶
# 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.
"""
cur = root
while cur:
if cur.left:
pre = cur.left
while pre.right:
pre = pre.right
pre.right = cur.right
cur.right = cur.left
cur.left = None
cur = cur.right