Leetcode 951: Flip Equivalent Binary Trees
You are given the roots of two binary trees. A flip operation consists of choosing any node and swapping its left and right child subtrees. A tree X is flip equivalent to tree Y if and only if we can make tree X equal to tree Y by performing some flip operations on X. Your task is to determine if the two trees are flip equivalent.
Problem
Approach
Steps
Complexity
Input: The input consists of the roots of two binary trees, represented as TreeNode objects.
Example: Input: root1 = [3,5,6,7,8], root2 = [3,8,5,6,7]
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].
Output: The output should be a boolean value indicating whether the two trees are flip equivalent.
Example: Output: true
Constraints:
• The output will be a boolean: true if the trees are flip equivalent, false otherwise.
Goal: To determine if two binary trees are flip equivalent, we need to recursively compare both trees while considering the possibility of flipping subtrees.
Steps:
• 1. Compare the root values of both trees. If they are not equal, return false.
• 2. Recursively check if both left and right subtrees are flip equivalent, either in the original order or after flipping them.
• 3. If either of these conditions hold true, return true; otherwise, return false.
Goal: The algorithm must be efficient to handle trees with up to 100 nodes.
Steps:
• Each tree can have at most 100 nodes.
• Each node's value is unique and lies within the range [0, 99].
Assumptions:
• The input trees are binary trees with unique node values.
• Input: Input: root1 = [1,2,3,4,5,6], root2 = [1,3,2,5,6,4]
• Explanation: Both trees are flip equivalent because we can perform flip operations at nodes with values 1 and 3 to make the structures of the two trees identical.
• Input: Input: root1 = [1,2,3], root2 = [1,3,2]
• Explanation: The trees are flip equivalent because we can swap the left and right subtrees of node 1.
Approach: The approach to solving this problem involves a recursive check on the binary trees. At each node, we will compare the left and right subtrees, considering both the possibility of no flip and a flip operation.
Observations:
• This problem can be solved by comparing trees recursively while considering both the direct and flipped structures.
• The recursive approach will allow us to break down the problem by comparing smaller subtrees at each level.
Steps:
• 1. Define a recursive function that checks if two subtrees are flip equivalent.
• 2. In the recursive function, check if the current nodes are equal and then recursively compare the left and right children in both original and flipped orders.
• 3. Return true if the subtrees are flip equivalent, otherwise false.
Empty Inputs:
• If both trees are empty, return true since two empty trees are trivially flip equivalent.
Large Inputs:
• For large trees with up to 100 nodes, the solution must work efficiently within the problem's constraints.
Special Values:
• If one tree is empty and the other is not, the trees are not flip equivalent.
Constraints:
• The solution must handle trees with at most 100 nodes efficiently.
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if(root1 == NULL || root2 == NULL)
return root1 == root2;
return (root1->val == root2->val) &&
(flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right) ||
flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left));
}
1 : Function Declaration
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
The function `flipEquiv` takes two binary tree roots, `root1` and `root2`, and returns `true` if the two trees are flip equivalent, otherwise `false`.
2 : Base Case Check
if(root1 == NULL || root2 == NULL)
Check if either of the trees is empty. If one is empty and the other is not, return `false`. If both are `NULL`, return `true`.
3 : Base Case Result
return root1 == root2;
Return `true` if both trees are `NULL` or `false` if one is `NULL` and the other is not.
4 : Recursive Check Start
Proceed to recursively compare the subtrees of both trees, considering both possible scenarios (subtrees are in the same order or flipped).
5 : Value Comparison
return (root1->val == root2->val) &&
First, check if the values at the current nodes of `root1` and `root2` are equal. If they are not equal, the trees cannot be flip equivalent.
6 : Recursive Left-Right Comparison
(flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right) ||
Recursively check if the left subtree of `root1` is flip equivalent to the left subtree of `root2`, and if the right subtrees match as well.
7 : Recursive Flipped Left-Right Comparison
flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left));
Alternatively, check if the left subtree of `root1` is flip equivalent to the right subtree of `root2`, and vice versa, indicating a flip of subtrees.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear, O(n), where n is the number of nodes in the tree. This is because we traverse each node once.
Best Case: O(h)
Worst Case: O(h)
Description: The space complexity is O(h), where h is the height of the tree, due to the recursion stack.
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