Description
https://leetcode.com/problems/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
TreeNodeclass where therightchild pointer points to the next node in the list and theleftchild 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)?
Explanation
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.
"""
self.helper(root)
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).