Leetcode #971: Flip Binary Tree To Match Preorder Traversal
In this guide, we solve Leetcode #971 Flip Binary Tree To Match Preorder Traversal 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
You are given the root of a binary tree with n nodes, where each node is uniquely assigned a value from 1 to n. You are also given a sequence of n values voyage, which is the desired pre-order traversal of the binary tree.
Quick Facts
- Difficulty: Medium
- Premium: No
- 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 = [1,2], voyage = [2,1]
Output: [-1]
Explanation: It is impossible to flip the nodes such that the pre-order traversal matches voyage.
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 flipMatchVoyage(self, root: Optional[TreeNode], voyage: List[int]) -> List[int]:
def dfs(root):
nonlocal i, ok
if root is None or not ok:
return
if root.val != voyage[i]:
ok = False
return
i += 1
if root.left is None or root.left.val == voyage[i]:
dfs(root.left)
dfs(root.right)
else:
ans.append(root.val)
dfs(root.right)
dfs(root.left)
ans = []
i = 0
ok = True
dfs(root)
return ans if ok else [-1]
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
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.