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0114.py
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# Source: https://leetcode.com/problems/flatten-binary-tree-to-linked-list
# Title: Flatten Binary Tree to Linked List
# Difficulty: Medium
# Author: Mu Yang <http://muyang.pro>
################################################################################################################################
# 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:
#
# https://assets.leetcode.com/uploads/2021/01/14/flaten.jpg
#
# 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
#
################################################################################################################################
# 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 __init__(self):
self.prev = None
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not root:
return
self.flatten(root.right)
self.flatten(root.left)
root.right = self.prev
root.left = None
self.prev = root
################################################################################################################################
class Solution2:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
self.flattenInner(root)
def flattenInner(self, root):
if not root:
return None
# Flatten right
right_end = self.flattenInner(root.right)
# Flatten left
left_end = self.flattenInner(root.left)
# Move right to end of left, move left to right
if left_end:
left_end.right = root.right
root.right = root.left
root.left = None
return right_end or left_end or root