#951

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Flip Equivalent Binary Trees

Determine if two binary trees are flip equivalent by recursively swapping subtrees.

Problem Statement

In this problem, you are given two binary trees, root1 and root2. The task is to determine whether the two trees are flip equivalent. A flip operation allows you to swap the left and right child subtrees of any node in the tree. Trees are flip equivalent if you can transform one tree into the other by performing zero or more flip operations.

To solve the problem, you must check if the trees can be made identical after applying flip operations. This involves comparing the structure and node values of both trees after accounting for any potential flips.

Examples

Example 1

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]

Output: true

We flipped at nodes with values 1, 3, and 5.

Example 2

Input: root1 = [], root2 = []

Output: true

Example details omitted.

Example 3

Input: root1 = [], root2 = [1]

Output: false

Example details omitted.

Constraints

The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].

Solution Approach

Depth-First Search (DFS) Traversal

To solve the problem, a DFS traversal of both trees is performed recursively. During each step, you check if the current node in both trees is equal. If they are, you proceed to check both possible subtree configurations (with and without a flip).

State Tracking with Recursion

While traversing the trees, you track the current state by recursively checking both flipped and non-flipped subtrees. The recursive nature of the solution ensures that every possible configuration is checked to ensure equivalence.

Edge Case Handling

The solution needs to handle edge cases such as when one tree is empty and the other is not, or when both trees are empty. These cases should return false or true as per the problem's conditions.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(N)

The time complexity of the solution is O(N), where N is the number of nodes in the trees. This is because we traverse each node once. The space complexity is also O(N) due to the recursive call stack, which stores the state of the tree traversal at each level.

What Interviewers Usually Probe

The candidate should demonstrate knowledge of tree traversal techniques and recursion.
Candidates should show the ability to track state during the traversal to handle flipped subtrees.
Expect the candidate to handle edge cases efficiently, especially the scenario where one tree is empty and the other is not.

Common Pitfalls or Variants

Common pitfalls

Failure to account for flipped subtrees and incorrectly assuming only direct equality between nodes.
Not properly handling edge cases where one tree is empty or the trees have different structures.
Incorrectly managing recursion, leading to excessive stack usage or incorrect comparisons.

Follow-up variants

Handling unbalanced trees with different shapes and depths.
Optimizing space complexity by using an iterative approach or a non-recursive traversal.
Modifying the problem to check for flip equivalence with a limited number of allowed flips.

FAQ

What is a flip operation in the context of binary trees?

A flip operation in a binary tree involves swapping the left and right child subtrees of a given node. This can be done at any node to try and make the trees equivalent.

How do you determine if two binary trees are flip equivalent?

Two binary trees are flip equivalent if you can make them identical by swapping the left and right subtrees of certain nodes through recursive DFS traversal.

What is the time complexity of the solution?

The time complexity is O(N), where N is the number of nodes in the trees. This is because each node is visited once during the DFS traversal.

What happens if one tree is empty and the other is not?

If one tree is empty and the other is not, the trees are not flip equivalent, and the answer should be false.

What should be done when encountering unbalanced trees?

When dealing with unbalanced trees, ensure that the recursive approach accounts for the differing shapes, checking both flipped and non-flipped subtree configurations.

terminal

Solution

Solution 1

#### Python3

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

# 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 flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        def dfs(root1, root2):
            if root1 == root2 or (root1 is None and root2 is None):
                return True
            if root1 is None or root2 is None or root1.val != root2.val:
                return False
            return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
                dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
            )

        return dfs(root1, root2)

Continue Practicing

#938 range sum of bst #965 univalued binary tree #968 binary tree cameras