Leetcode #1666: Change the Root of a Binary Tree
In this guide, we solve Leetcode #1666 Change the Root of a Binary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given the root of a binary tree and a leaf node, reroot the tree so that the leaf is the new root. You can reroot the tree with the following steps for each node cur on the path starting from the leaf up to the root excluding the root: If cur has a left child, then that child becomes cur's right child.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Tree, Depth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [3,5,1,6,2,0,8,null,null,7,4], leaf = 7
Output: [7,2,null,5,4,3,6,null,null,null,1,null,null,0,8]
Python Solution
"""
# Definition for a Node.
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
self.parent = None
"""
class Solution:
def flipBinaryTree(self, root: "Node", leaf: "Node") -> "Node":
cur = leaf
p = cur.parent
while cur != root:
gp = p.parent
if cur.left:
cur.right = cur.left
cur.left = p
p.parent = cur
if p.left == cur:
p.left = None
elif p.right == cur:
p.right = None
cur = p
p = gp
leaf.parent = None
return leaf
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.