Problem Description

In this problem, we are given two binary trees represented by their root nodes root1 and root2. We need to determine if one tree can be transformed into the other tree through a series of flip operations. A flip operation consists of swapping the left and right children of a given node. The trees are considered flip equivalent if one can be transformed into the other by doing any number of flips, possibly zero. We should return true if the trees are flip equivalent and false otherwise.

Flowchart Walkthrough

Let's determine the appropriate algorithm using the Flowchart. We'll go through it step-by-step to analyze LeetCode problem 951, Flip Equivalent Binary Trees:

Is it a graph?

• Yes: Binary trees can be viewed as a type of graph with nodes (tree nodes) and edges (links between parent and children).

Is it a tree?

• Yes: The structure discussed in the problem is specifically described as a binary tree, which is a type of tree.

At this point, the flowchart leads us directly to the use of Depth-First Search (DFS), since from a tree node, we are given a direct indication to proceed with DFS:

DFS is particularly suited for tree operations as it allows us to explore each branch completely before moving to another branch, which is useful in determining if two trees are flip equivalent by recursively comparing each subtree.

Conclusion: Following the directions in the flowchart, Depth-First Search (DFS) is suggested for solving the problem of determining if two binary trees are flip equivalent.

Intuition

To solve this problem, we use a recursive approach that is a form of Depth-First Search (DFS). The main idea is that if two trees are flip equivalent, then their roots must have the same value, and either both pairs of the left and right subtrees are flip equivalent, or the left subtree of one tree is flip equivalent to the right subtree of the other tree and vice versa.

To implement this idea, we compare the current nodes of both trees:

1. If both nodes are None, then the trees are trivially flip equivalent.
2. If only one node is None or the values of the nodes differ, the trees are not flip equivalent.
3. If the nodes have the same value, we then recursively check their subtrees.

We have two cases for recursion to check for flip equivalence:

• Case 1: Left subtree of root1 is flip equivalent to left subtree of root2 AND right subtree of root1 is flip equivalent to right subtree of root2.
• Case 2: Left subtree of root1 is flip equivalent to right subtree of root2 AND right subtree of root1 is flip equivalent to left subtree of root2.

If either of these cases is true, we conclude that the trees rooted at root1 and root2 are flip equivalent.

This approach works well because it exploits the property that a tree is defined not just by its nodes and their values, but by the specific arrangement of these nodes. By checking all possible flip combinations of subtrees, we can effectively determine flip equivalence between the two given trees.

Learn more about Tree, Depth-First Search and Binary Tree patterns.

Solution Approach

The solution code implements a recursive function called dfs which stands for depth-first search. This function continuously dives deeper into the subtrees of both trees, simultaneously, for as long as it finds matching node values and structure that adhere to flip equivalence. Below are the aspects of the implementation explained:

• Base Cases: The first checks in the dfs function handle the base cases. If both nodes are equal, this means that we have reached equivalent leaves, or both nodes are None, which means the subtrees are empty and trivially flip equivalent. We thus return true. If one of the nodes is None while the other isn’t, or if the nodes’ values are not equal, the function immediately returns false indicating the trees are not flip equivalent at this level.

• Recursive Checks: The dfs function then makes two critical recursive calls representing the two possible scenarios to check for equivalence:

  1. The first scenario checks if the left subtree of root1 is equivalent to the left subtree of root2 and if the right subtree of root1 is equivalent to the right subtree of root2. This corresponds to the situation where no flips are necessary at the current level of the trees.

  2. The second scenario checks if the left subtree of root1 is equivalent to the right subtree of root2 and if the right subtree of root1 is equivalent to the left subtree of root2. Here, it matches the case where a flip at the current node would render the subtrees equivalent.

• Logical OR Operation: The or operation between the two checks is used to state that if either of these scenarios holds true, then the trees rooted at root1 and root2 are flip equivalent at the current level.

• Efficient Short-Circuiting: The use of the and within each scenario creates short-circuiting behavior. This means if the first check within a scenario returns false, the second check is not performed, saving unnecessary computation.

The dfs function is invoked with the root nodes of the two trees. The return value of this initial call will be the result for flip equivalence of the entire trees. This method effectively uses the call stack as a data structure to hold the state of each recursive call, allowing the algorithm to backtrace through the nodes of the trees and unwind the recursion with the correct answer.

No other explicit data structures are used, which implies that the space complexity of the solution is primarily dependent on the height of the recursion tree, while the time complexity depends on the size of the binary trees being compared.

Example Walkthrough

Let's take a small example of two binary trees to illustrate the solution approach:

Let tree A have the following structure:

     1
   /   \
  2     3
 / \   / \
4   5 6   7

And let tree B have a structure that is a flipped version of tree A:

     1
   /   \
  3     2
 / \   / \
7   6 5   4

We want to determine if tree B can be transformed into tree A via flip operations.

Application of the Solution Approach:

Step 1: We begin with the dfs function call on the root of tree A (which has the value 1), and the root of tree B (also value 1).

Step 2: Since both root nodes' values are identical and not None, we check their subtrees, implementing the two scenarios described in the solution approach.

Step 3: We check the left subtree of A (2) with the left subtree of B (3) and the right subtree of A (3) with the right subtree of B (2). Since the values don't match, we proceed to the next step without making any further recursive calls in this scenario.

Step 4: We then check the left subtree of A (2) with the right subtree of B (2) and the right subtree of A (3) with the left subtree of B (3). Here, as we first match the values, we need to delve deeper recursively.

Recursive Step: For each matched pair (2 with 2, 3 with 3), we proceed recursively:

• For the matched pair of nodes with value 2:

  • Compare left child of A's 2 (4) with right child of B's 2 (4).
  • Compare right child of A's 2 (5) with left child of B's 2 (5).
  • Both child pairs match in value; hence, recursion into them will return true.

• For the matched pair of nodes with value 3:

  • Compare left child of A's 3 (6) with the right child of B's 3 (6).
  • Compare right child of A's 3 (7) with the left child of B's 3 (7).
  • Again, both child pairs match in value and further recursive checks would yield true.

As all recursive checks return true, we can say that tree A is flip equivalent to tree B; hence, the initial call to dfs from the roots will return true, validating the flip equivalence of the two binary trees.

Solution Implementation

Time and Space Complexity

The code defines a recursive function dfs which checks if two binary trees are flip equivalent. The dfs function is called for each corresponding pair of nodes in the two trees. At each step, the code performs constant time operations before potentially making up to two more recursive calls.

The time complexity is O(N) where N is the smaller number of nodes in either root1 or root2. This is because the recursion stops at the leaf nodes or when a mismatch is found. In the worst case, every node in the smaller tree will be visited.

The space complexity is also O(N) due to the recursion stack. In the worst case, the recursion goes as deep as the height of the tree, which can be O(N) in the case of a skewed tree (a tree in which every internal node has only one child). For a balanced tree, the space complexity will be O(log N) because the height of the tree would be logarithmic relative to the number of nodes.

Learn more about how to find time and space complexity quickly using problem constraints.