For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.
A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.
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 Explanation: We flipped at nodes with values 1, 3, and 5.
Example 2:
Input: root1 = [], root2 = [] Output: true
Example 3:
Input: root1 = [], root2 = [1] Output: false
Constraints:
[0, 100].[0, 99].The key idea for Flip Equivalent Binary Trees is to determine whether two binary trees can become identical after any number of flip operations. A flip operation swaps the left and right children of a node. This naturally suggests a Depth-First Search (DFS) comparison of the two trees.
Start by checking simple base cases such as both nodes being null or having different values. If the values match, recursively verify two possibilities: either the left subtree of the first tree matches the left subtree of the second and the right matches the right, or the subtrees match in a flipped manner (left with right and right with left). If either configuration returns true, the trees are flip equivalent.
This recursive strategy explores structural equivalence while allowing swaps at any level. The algorithm visits each node once in the best scenario, making it efficient for typical binary tree sizes.
Time Complexity: O(n) where n is the number of nodes. Space Complexity: O(h) due to recursion stack, where h is the tree height.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| DFS Recursive Tree Comparison | O(n) | O(h) |
NeetCode
This approach uses a recursive function to check if both trees are identical or one is a flipped version of the other. For each pair of nodes, we check if subtree structures match directly or with a flip.
Time Complexity: O(N), where N is the total number of nodes, since we visit each node once.
Space Complexity: O(H), where H is the height of the tree, due to the recursion stack.
1#include <stdbool.h>
2
3struct TreeNode {
4 int val;
5 struct TreeNode *left;
6 struct TreeNode *right;
7};
8
9bool flipEquiv(struct TreeNode* root1, struct TreeNode* root2) {
10 if (!root1 && !root2) return true;
11 if (!root1 || !root2 || root1->val != root2->val) return false;
12
13 return (flipEquiv(root1->left, root2->left) &&
14 flipEquiv(root1->right, root2->right)) ||
15 (flipEquiv(root1->left, root2->right) &&
16 flipEquiv(root1->right, root2->left));
17}In this C implementation, a recursive function flipEquiv is called for each node. It first checks base cases, including if both nodes are null (return true), or if one is null and the other isn't (return false). Then it checks if the current nodes are equal in value and recursively checks for their children in both non-flipped and flipped scenarios.
In this approach, instead of recursion, we use a stack for depth-first search. It proceeds similarly by evaluating if nodes are flip equivalent, but uses an explicit stack rather than the call stack.
Time Complexity: O(N).
Space Complexity: O(H), similar to the recursive approach.
1class TreeNode:
2 def __init__(self, x)
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Yes, variations of tree comparison and structural equivalence problems are common in FAANG interviews. This problem tests recursion, tree traversal, and the ability to reason about structural transformations.
DFS works well because the problem requires comparing nodes and their subtrees recursively. It allows the algorithm to explore both flipped and non-flipped subtree configurations at each node efficiently.
The optimal approach uses a recursive Depth-First Search to compare corresponding nodes in both trees. At each node, the algorithm checks both non-flipped and flipped subtree combinations. If either configuration matches, the trees are considered flip equivalent.
A binary tree structure combined with recursion is the most suitable choice. DFS traversal allows you to efficiently compare node values and subtree structures while accounting for possible flips.
In this Python implementation, we use a stack to iteratively check node pairs. We continue popping nodes to compare for equivalency in both standard and flipped orientations until the stack is empty.